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6.6: Divide Polynomials


Learning Objectives

By the end of this section, you will be able to:

  • Divide a polynomial by a monomial
  • Divide a polynomial by a binomial

Note

Before you get started, take this readiness quiz.

  1. Add: (dfrac{3}{d}+dfrac{x}{d})
    If you missed this problem, review Exercise 1.7.1.
  2. Simplify: (dfrac{30 x y^{3}}{5 x y})
    If you missed this problem, review Exercise 6.5.37.
  3. Combine like terms: (8 a^{2}+12 a+1+3 a^{2}-5 a+4)
    If you missed this problem, review Exercise 1.3.37.

Divide a Polynomial by a Monomial

In the last section, you learned how to divide a monomial by a monomial. As you continue to build up your knowledge of polynomials the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition.

(egin{array}{ll}{ ext { The sum, }} & {dfrac{y}{5}+dfrac{2}{5}} { ext { simplifies to }} & {dfrac{y+2}{5}}end{array})

Now we will do this in reverse to split a single fraction into separate fractions.

We’ll state the fraction addition property here just as you learned it and in reverse.

FRACTION ADDITION

If a,b, and c are numbers where (c eq 0), then

[dfrac{a}{c}+dfrac{b}{c}=dfrac{a+b}{c} quad ext { and } quad dfrac{a+b}{c}=dfrac{a}{c}+dfrac{b}{c}]

We use the form on the left to add fractions and we use the form on the right to divide a polynomial by a monomial.

(egin{array}{ll}{ ext { For example, }} & {dfrac{y+2}{5}} { ext { can be written }} & {dfrac{y}{5}+dfrac{2}{5}}end{array})

We use this form of fraction addition to divide polynomials by monomials.

DIVISION OF A POLYNOMIAL BY A MONOMIAL

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Exercise (PageIndex{1})

Find the quotient:(dfrac{7 y^{2}+21}{7})

Answer

(egin{array}{ll} & dfrac{7 y^{2}+21}{7} ext{Divide each term of the numerator by the denominator.} & dfrac{7 y^{2}}{7}+dfrac{21}{7} ext {Simplify each fraction. } & y^{2}+3 end{array})

Exercise (PageIndex{2})

Find the quotient: (dfrac{8 z^{2}+24}{4})

Answer

(2 z^{2}+6)

Exercise (PageIndex{3})

Find the quotient:(dfrac{18 z^{2}-27}{9})

Answer

(2 z^{2}-3)

Remember that division can be represented as a fraction. When you are asked to divide a polynomial by a monomial and it is not already in fraction form, write a fraction with the polynomial in the numerator and the monomial in the denominator.

Exercise (PageIndex{4})

Find the quotient: (left(18 x^{3}-36 x^{2} ight) div 6 x)

Answer

(egin{array}{ll} & left(18 x^{3}-36 x^{2} ight) div 6 x ext { Rewrite as a fraction. } & dfrac{18 x^{3}-36 x^{2}}{6 x} ext { Divide each term of the numerator by the denominator. }& dfrac{18 x^{3}}{6 x}-dfrac{36 x^{2}}{6 x} ext { Simplify. } &3 x^{2}-6 xend{array})

Exercise (PageIndex{5})

Find the quotient: (left(27 b^{3}-33 b^{2} ight) div 3 b)

Answer

(9 b^{2}-11 b)

Exercise (PageIndex{6})

Find the quotient: (left(25 y^{3}-55 y^{2} ight) div 5 y)

Answer

(5 y^{2}-11 y)

When we divide by a negative, we must be extra careful with the signs.

Exercise (PageIndex{7})

Find the quotient: (dfrac{12 d^{2}-16 d}{-4})

Answer

(egin{array}{ll} &dfrac{12 d^{2}-16 d}{-4} ext { Divide each term of the numerator by the denominator. }& dfrac{18 x^{3}-36 x^{2}}{6 x} ext { Simplify. Remember, subtracting a negative is like adding a positive! }& -3 d^{2}+4 dend{array})

Exercise (PageIndex{8})

Find the quotient: (dfrac{25 y^{2}-15 y}{-5})

Answer

(-5 y^{2}+3 y)

Exercise (PageIndex{9})

Find the quotient: (dfrac{42 b^{2}-18 b}{-6})

Answer

(-7 b^{2}+3 b)

Exercise (PageIndex{10})

Find the quotient: (dfrac{105 y^{5}+75 y^{3}}{5 y^{2}})

Answer

(egin{array}{ll} &dfrac{105 y^{5}+75 y^{3}}{5 y^{2}} ext { Separate the terms. }& dfrac{105 y^{5}}{5 y^{2}}+dfrac{75 y^{3}}{5 y^{2}} ext { Simplify. }& 21 y^{3}+15 yend{array})

Exercise (PageIndex{11})

Find the quotient: (dfrac{60 d^{7}+24 d^{5}}{4 d^{3}})

Answer

(15 d^{4}+6 d^{2})

Exercise (PageIndex{12})

Find the quotient: (dfrac{216 p^{7}-48 p^{5}}{6 p^{3}})

Answer

(36 p^{4}-8 p^{2})

Exercise (PageIndex{13})

Find the quotient: (left(15 x^{3} y-35 x y^{2} ight) div(-5 x y))

Answer

(egin{array}{ll} &left(15 x^{3} y-35 x y^{2} ight) div(-5 x y) ext { Rewrite as a fraction. }& dfrac{15 x^{3} y-35 x y^{2}}{-5 x y} ext { Separate the terms. Be careful with the signs! }& dfrac{15 x^{3} y}{-5 x y}-dfrac{35 x y^{2}}{-5 x y} ext { Simplify. } & -3 x^{2}+7 yend{array})

Exercise (PageIndex{14})

Find the quotient: (left(32 a^{2} b-16 a b^{2} ight) div(-8 a b))

Answer

(-4 a+2 b)

Exercise (PageIndex{15})

Find the quotient: (left(-48 a^{8} b^{4}-36 a^{6} b^{5} ight) divleft(-6 a^{3} b^{3} ight))

Answer

(8 a^{5} b+6 a^{3} b^{2})

Exercise (PageIndex{13})

Find the quotient: (dfrac{36 x^{3} y^{2}+27 x^{2} y^{2}-9 x^{2} y^{3}}{9 x^{2} y})

Answer

(egin{array}{ll} &dfrac{36 x^{3} y^{2}+27 x^{2} y^{2}-9 x^{2} y^{3}}{9 x^{2} y} ext { Separate the terms. }& dfrac{36 x^{3} y^{2}}{9 x^{2} y}+dfrac{27 x^{2} y^{2}}{9 x^{2} y}-dfrac{9 x^{2} y^{3}}{9 x^{2} y} ext { Simplify. } & 4 x y+3 y-y^{2}end{array})

Exercise (PageIndex{14})

Find the quotient: (dfrac{40 x^{3} y^{2}+24 x^{2} y^{2}-16 x^{2} y^{3}}{8 x^{2} y})

Answer

(5 x y+3 y-2 y^{2})

Exercise (PageIndex{15})

Find the quotient: (dfrac{35 a^{4} b^{2}+14 a^{4} b^{3}-42 a^{2} b^{4}}{7 a^{2} b^{2}})

Answer

(5 a^{2}+2 a^{2} b-6 b^{2})

Exercise (PageIndex{16})

Find the quotient: (dfrac{10 x^{2}+5 x-20}{5 x})

Answer

(egin{array}{ll}&dfrac{10 x^{2}+5 x-20}{5x} ext { Separate the terms. }& dfrac{10 x^{2}}{5 x}+dfrac{5 x}{5 x}-dfrac{20}{5 x} ext { Simplify. } &2 x+1-dfrac{4}{x}end{array})

Exercise (PageIndex{17})

Find the quotient: (dfrac{18 c^{2}+6 c-9}{6 c})

Answer

(3 c+1-dfrac{3}{2 c})

Exercise (PageIndex{18})

Find the quotient: (dfrac{10 d^{2}-5 d-2}{5 d})

Answer

(2 d-1-dfrac{2}{5 d})

Divide a Polynomial by a Binomial

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

We write the long division
We divide the first two digits, 87, by 25.
We multiply 3 times 25 and write the product under the 87.
Now we subtract 75 from 87.
Then we bring down the third digit of the dividend, 5.
Repeat the process, dividing 25 into 125.

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

[egin{array}{l}{35 cdot 25} {875}checkmarkend{array}]

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Exercise (PageIndex{20})

Find the quotient: (left(y^{2}+10 y+21 ight) div(y+3))

Answer

y+7

Exercise (PageIndex{21})

Find the quotient: (left(m^{2}+9 m+20 ight) div(m+4))

Answer

m+5

When the divisor has subtraction sign, we must be extra careful when we multiply the partial quotient and then subtract. It may be safer to show that we change the signs and then add.

Exercise (PageIndex{23})

Find the quotient: (left(2 x^{2}-3 x-20 ight) div(x-4))

Answer

2x+5

Exercise (PageIndex{24})

Find the quotient: (left(3 x^{2}-16 x-12 ight) div(x-6))

Answer

3x+2

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In Exercise (PageIndex{25}), we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Exercise (PageIndex{26})

Find the quotient: (left(x^{3}+5 x^{2}+8 x+6 ight) div(x+2))

Answer

(x^{2}+3 x+2+dfrac{2}{x+2})

Exercise (PageIndex{27})

Find the quotient: (left(2 x^{3}+8 x^{2}+x-8 ight) div(x+1))

Answer

(2 x^{2}+6 x-5-dfrac{3}{x+1})

Look back at the dividends in Example, Example, and Example. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in Example will be (x^{4}-x^{2}+5 x-2). It is missing an (x^{3}) term. We will add in (0x^{3}) as a placeholder.

Exercise (PageIndex{28})

Find the quotient: (left(x^{4}-x^{2}+5 x-2 ight) div(x+2))

Answer

Notice that there is no (x^{3}) term in the dividend. We will add (0x^{3}) as a placeholder.

Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms.
Divide x4 by x.
Put the answer, x3, in the quotient over the x3 term.
Multiply x3 times x + 2. Line up the like terms.
Subtract and then bring down the next term.
Divide −2x3 by x.
Put the answer, −2x2, in the quotient over the x2 term.
Multiply −2x2 times x + 1. Line up the like terms.
Subtract and bring down the next term.
Divide 3x2 by x.
Put the answer, 3x, in the quotient over the x term.
Multiply 3x times x + 1. Line up the like terms.
Subtract and bring down the next term.
Divide −x by x.
Put the answer, −1, in the quotient over the constant term.
Multiply −1 times x + 1. Line up the like terms.
Change the signs, add.
The result should be (x^{4}-x^{2}+5 x-2)

Exercise (PageIndex{29})

Find the quotient: (left(x^{3}+3 x+14 ight) div(x+2))

Answer

(x^{2}-2 x+7)

Exercise (PageIndex{30})

Find the quotient: (left(x^{4}-3 x^{3}-1000 ight) div(x+5))

Answer

(x^{3}-8 x^{2}+40 x-200)

In Exercise (PageIndex{31}), we will divide by (2a−3). As we divide we will have to consider the constants as well as the variables.

Exercise (PageIndex{31})

Find the quotient: (left(8 a^{3}+27 ight) div(2 a+3))

Answer

This time we will show the division all in one step. We need to add two placeholders in order to divide.

To check, multiply ((2 a+3)left(4 a^{2}-6 a+9 ight))

The result should be (8 a^{3}+27)

Exercise (PageIndex{32})

Find the quotient: (left(x^{3}-64 ight) div(x-4))

Answer

(x^{2}+4 x+16)

Exercise (PageIndex{33})

Find the quotient: (left(125 x^{3}-8 ight) div(5 x-2))

Answer

(25 x^{2}+10 x+4)

Note

Access these online resources for additional instruction and practice with dividing polynomials:

  • Divide a Polynomial by a Monomial
  • Divide a Polynomial by a Monomial 2
  • Divide Polynomial by Binomial

Key Concepts

  • Fraction Addition
    • If a,b, and c are numbers where (c eq 0), then
      (dfrac{a}{c}+dfrac{b}{c}=dfrac{a+b}{c}) and (dfrac{a+b}{c}=dfrac{a}{c}+dfrac{b}{c})
  • Division of a Polynomial by a Monomial
    • To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Division of Polynomials: Definition, Method, and Examples

Division of Polynomials: Polynomials are algebraic expressions consisting of variables and constants such that the exponent on the variables is a whole number. We can perform arithmetic operations such as addition, subtraction, multiplication, and division with polynomials.

Polynomial is derived from the Greek word. Poly means many and nomial means terms, so together, we can call a polynomial as many terms. So a polynomial has one or more than one term. The division of polynomials follows the same rules that we use to follow in the division of integers. This article details polynomials, their degree, types, and how to carry out the division of polynomials.


Long Division Of Polynomials

In these lessons, we will learn how to divide a polynomial with another polynomial using long division.

Division of one polynomial by another requires a process somewhat like long division in arithmetic. Now, however, we will use polynomials instead of just numerical values.

The following diagram shows an example of polynomial division using long division. Scroll down the page for more examples and solutions on polynomial division.


Example:
Evaluate (x 2 + 10x + 21) ÷ (x + 7) using long division.

Solution:
(x 2 + 10x + 21) is called the dividend and (x + 7) is called the divisor.

Step 1: Divide the first term of the dividend with the first term of the divisor and write the result as the first term of the quotient.

Step 2: Multiply that term with the divisor.

Step 3: Subtract and write the result to be used as the new dividend.

Step 4: Divide the first term of this new dividend by the first term of the divisor and write the result as the second term of the quotient.

Step 5: Multiply that term and the divisor and write the result under the new dividends.

Step 6: Subtract to get the remainder.

Note that it also possible that the remainder of a polynomial division may not be zero.

Example:
Evaluate (23y 2 + 9 + 20y 3 – 13y) ÷ (2 + 5y 2 – 3y)

You may want to look at the lesson on synthetic division, a simplified form of long division.

Dividing Polynomials using Long Division
When dividing polynomials, we can use either long division or synthetic division to arrive at an answer. Using long division, dividing polynomials is easy. We simply write the fraction in long division form by putting the divisor outside of the bracket and the divided inside the bracket. After the polynomial division is set up, we follow the same process as long division with numbers.

Example:
(3x 3 - 4x 2 + 2x - 1) ÷ (x + 1)

Dividing Polynomials by Binomials

Example:
(x 2 + 5x + 6) ÷ (x + 1)

Dividing Polynomials

Example:
(x 2 + 9x + 20) ÷ (x + 5)
(6x 2 + 7x - 20) ÷ (2x + 5)
(6x 4 - 30x 2 + 24) ÷ (2x 2 - 8)
(3x 5 + 4x 3 - 5x + 8) ÷ (x 2 + 3)
(x 5 + 2x 4 + x 3 - x 2 - 22x + 15) ÷ (x 2 + 2x - 3)

How to divide polynomials using long division?

How to do Long Division with Polynomials with remainder?

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DIVIDING POLYNOMIALS WITH MISSING TERMS

For example, the highest degree of the polynomial is 3, then the next term of the dividend must be the square term and so on. In this, if we don't have square term, we have to write 0 instead of this and we can write the next terms.

Let us see some example problems based on the above concept.

Do the following division : 

The degree of the given polynomial is 3 then we must have square term. But here we don't have square term, so we have to replace it by 0.

In the first step, we have to divide the first term of the dividend by the first term of the divisor.

If we divide  a 3 ਋y a, we will get a 2 . We have to write  a 2 ਊt the top and multiply each term of the dividend by  a 2 .

By subtracting  a 3  -  2 a 2 ਏrom  ( a 3  + 8a - 24), we get 

Now we have to divide 2 a 2 ਋y a, so we will get 2a.

  • By multiplying this 2a by (a - 2), we get  2 a 2  - 4a.
  • By subtracting  2 a 2 - 4a from  2 a 2  - 8a - 24, we get 12a - 24

Divide 12a ਋y a, so we get 12.

  • By multiplying this 12 by (a - 2), we get 12a - 24.
  • By subtracting 12a - 24 from 12a - 24, we get 0.

Do the following division :

The degree of the given polynomial is 3. But here we don't have x term, so we have to replace it by 0.

In the first step, we have to divide the first term of the dividend by the first term of the divisor.

If we divide  4 x 3 ਋y 2x, we will get 2 x 2 .

  • Write this 2x 2  at the top.
  • Multiply 2 x 2  by 2x.  So we get 4x 3 .
  • Subtract  4x 3 ਏrom  4x 3  + 2 x 2 .

Now we have to divide the first term of the dividend 2 x 2 ਋y 2x, so we get x.

  • By multiplying x by 2x, we get  2x 2 .
  • Subtracting  2x 2 ਏorm  2x 2  we get 0
  • Bring down the next term, we get -5

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Modular Arithmetic

That is where the modular arithmetic comes into play. The idea of modular arithmetic is following: instead of having an infinite set of numbers we declare that we select only first n natural numbers, i.e., 0, 1, …, n – 1, to work with, and if any given integer falls out of this range, we “wrap” it around. For example, let us choose six first numbers. To illustrate this, consider a circle with six ticks of equal units this is our range (usually referred to as finite field).

Now let us see where the number eight will land. As an analogy, we can think of it as a rope, the length of which is eight units:

If we attach the rope to the beginning of the circle

and start wrapping the rope around it, after one rotation we still have a portion of the rope left:

Therefore if we continue the process, the rope will end right at the tick #2.

It is the result of the modulo operation. No matter how long the rope is it will always stop at one of the circle’s ticks. Therefore the modulo operation will keep it in certain bounds (in this case from 0 to 5). The 15-units rope will stop at 3, i.e., 6 + 6 + 3 (two full circles with 3-units leftover). The negative numbers work the same way, and the only difference is that we wrap it in the opposite direction, for –8 the result will be 4.

Moreover, we can perform arithmetic operations, and the result will always be in the scope of n numbers. We will use the notation “mod n” for now on to denote the range of numbers. For example:

Furthermore, the most important property is that the order of operations does not matter, e.g., we can perform all operations first and then apply modulo or apply modulo after every operation. For example (2 × 4 – 1) × 3 = 3 (mod 6) is equivalent to:

So why on earth is that helpful? It turns out that if we use modulo arithmetic, having a result of operation it is non-trivial to go back to the original numbers because many different combinations will have the same result:

Without the modular arithmetic, the size of the result gives a clue to its solution. This piece of information is hidden otherwise, while common arithmetic properties are preserved.


Contents

Let Fq = GF(q) be the finite field of characteristic p , that is, the field having q elements where q = p e for some prime p . A polynomial f with coefficients in Fq (symbolically written as fFq[x] ) is a permutation polynomial of Fq if the function from Fq to itself defined by c ↦ f ( c ) is a permutation of Fq . [3]

Due to the finiteness of Fq , this definition can be expressed in several equivalent ways: [4]

  • the function c ↦ f ( c ) is onto (surjective)
  • the function c ↦ f ( c ) is one-to-one (injective)
  • f(x) = a has a solution in Fq for each a in Fq
  • f(x) = a has a unique solution in Fq for each a in Fq .

A characterization of which polynomials are permutation polynomials is given by

(Hermite's Criterion) [5] [6] fFq[x] is a permutation polynomial of Fq if and only if the following two conditions hold:

  1. f has exactly one root in Fq
  2. for each integer t with 1 ≤ tq − 2 and t ≢ 0 ( mod p ) >> , the reduction of f(x) t mod (xqx) has degree ≤ q − 2 .

If f(x) is a permutation polynomial defined over the finite field GF(q) , then so is g(x) = a f(x + b) + c for all a ≠ 0, b and c in GF(q) . The permutation polynomial g(x) is in normalized form if a, b and c are chosen so that g(x) is monic, g(0) = 0 and (provided the characteristic p does not divide the degree n of the polynomial) the coefficient of x n-1 is 0.

There are many open questions concerning permutation polynomials defined over finite fields (see Lidl & Mullen (1988) and Lidl & Mullen (1993)).

Small degree Edit

Hermite's criterion is computationally intensive and can be difficult to use in making theoretical conclusions. However, Dickson was able to use it to find all permutation polynomials of degree at most five over all finite fields. These results are: [7] [6]

Normalized Permutation Polynomial of Fq q
x any q
x 2 > q ≡ 0 ( mod 2 ) >>
x 3 > q ≢ 1 ( mod 3 ) >>
x 3 − a x -ax> ( a not a square) q ≡ 0 ( mod 3 ) >>
x 4 ± 3 x pm 3x> q = 7
x 4 + a 1 x 2 + a 2 x +a_<1>x^<2>+a_<2>x> (if its only root in Fq is 0) q ≡ 0 ( mod 2 ) >>
x 5 > q ≢ 1 ( mod 5 ) >>
x 5 − a x -ax> ( a not a fourth power) q ≡ 0 ( mod 5 ) >>
x 5 + a x ( a 2 = 2 ) +ax,(a^<2>=2)> q = 9
x 5 ± 2 x 2 pm 2x^<2>> q = 7
x 5 + a x 3 ± x 2 + 3 a 2 x +ax^<3>pm x^<2>+3a^<2>x> ( a not a square) q = 7
x 5 + a x 3 + 5 − 1 a 2 x +ax^<3>+5^<-1>a^<2>x> ( a arbitrary) q ≡ ± 2 ( mod 5 ) >>
x 5 + a x 3 + 3 a 2 x +ax^<3>+3a^<2>x> ( a not a square) q = 13
x 5 − 2 a x 3 + a 2 x -2ax^<3>+a^<2>x> ( a not a square) q ≡ 0 ( mod 5 ) >>

A list of all monic permutation polynomials of degree six in normalized form can be found in Shallue & Wanless (2013). [8]

Some classes of permutation polynomials Edit

Beyond the above examples, the following list, while not exhaustive, contains almost all of the known major classes of permutation polynomials over finite fields. [9]

  • xn permutes GF(q) if and only if n and q − 1 are coprime (notationally, (n, q − 1) = 1 ). [10]
  • If a is in GF(q) and n ≥ 1 then the Dickson polynomial (of the first kind) Dn(x,a) is defined by

These can also be obtained from the recursion

If a ≠ 0 and n > 1 then Dn(x, a) permutes GF(q) if and only if (n, q 2 − 1) = 1 . [11] If a = 0 then Dn(x, 0) = x n and the previous result holds.

The linearized polynomials that are permutation polynomials over GF(q r ) form a group under the operation of composition modulo x q r − x >-x> , which is known as the Betti-Mathieu group, isomorphic to the general linear group GL(r, Fq) . [12]

  • If g(x) is in the polynomial ring Fq[x] and g(xs ) has no nonzero root in GF(q) when s divides q − 1 , and r > 1 is relatively prime (coprime) to q − 1 , then xr (g(xs )) (q - 1)/s permutes GF(q) . [6]
  • Only a few other specific classes of permutation polynomials over GF(q) have been characterized. Two of these, for example, are:

Exceptional polynomials Edit

An exceptional polynomial over GF(q) is a polynomial in Fq[x] which is a permutation polynomial on GF(q m ) for infinitely many m . [13]

A permutation polynomial over GF(q) of degree at most q 1/4 is exceptional over GF(q) . [14]

Every permutation of GF(q) is induced by an exceptional polynomial. [14]

If a polynomial with integer coefficients (i.e., in ℤ[x] ) is a permutation polynomial over GF(p) for infinitely many primes p , then it is the composition of linear and Dickson polynomials. [15] (See Schur's conjecture below).

In finite geometry coordinate descriptions of certain point sets can provide examples of permutation polynomials of higher degree. In particular, the points forming an oval in a finite projective plane, PG(2,q) with q a power of 2, can be coordinatized in such a way that the relationship between the coordinates is given by an o-polynomial, which is a special type of permutation polynomial over the finite field GF(q) .

The problem of testing whether a given polynomial over a finite field is a permutation polynomial can be solved in polynomial time. [16]

For the finite ring Z/nZ one can construct quadratic permutation polynomials. Actually it is possible if and only if n is divisible by p 2 for some prime number p. The construction is surprisingly simple, nevertheless it can produce permutations with certain good properties. That is why it has been used in the interleaver component of turbo codes in 3GPP Long Term Evolution mobile telecommunication standard (see 3GPP technical specification 36.212 [18] e.g. page 14 in version 8.8.0).

Simple examples Edit

g(3)=1> , so the polynomial defines the permutation

g(7)=1> , so the polynomial defines the permutation

Rings Z/p k Z Edit

Lemma: for k=1 (i.e. Z/pZ) such polynomial defines a permutation only in the case a=0 and b not equal to zero. So the polynomial is not quadratic, but linear.

Lemma: for k>1, p>2 (Z/p k Z) such polynomial defines a permutation if and only if a ≡ 0 ( mod p ) >> and b ≢ 0 ( mod p ) >> .

Rings Z/nZ Edit

As a corollary one can construct plenty quadratic permutation polynomials using the following simple construction. Consider n = p 1 k 1 p 2 k 2 . . . p l k l ^>p_<2>^>. p_^<>>> , assume that k1 >1.

p_<1>> , but a ≠ 0 m o d p 1 k 1

p_<1>^>> assume that a = 0 m o d p i k i

p_^<>>> ,i>1. And assume that b ≠ 0 m o d p i

To see this we observe that for all primes pi,i>1, the reduction of this quadratic polynomial modulo pi is actually linear polynomial and hence is permutation by trivial reason. For the first prime number we should use the lemma discussed previously to see that it defines the permutation.

For example, consider Z/12Z and polynomial 6 x 2 + x +x> . It defines a permutation

A polynomial g(x) for the ring Z/p k Z is a permutation polynomial if and only if it permutes the finite field Z/pZ and g ′ ( x ) ≠ 0 m o d p

p> for all x in Z/p k Z, where g′(x) is the formal derivative of g(x). [19]

Let K be an algebraic number field with R the ring of integers. The term "Schur's conjecture" refers to the assertion that, if a polynomial f defined over K is a permutation polynomial on R/P for infinitely many prime ideals P, then f is the composition of Dickson polynomials, degree-one polynomials, and polynomials of the form x k . In fact, Schur did not make any conjecture in this direction. The notion that he did is due to Fried, [20] who gave a flawed proof of a false version of the result. Correct proofs have been given by Turnwald [21] and Müller. [22]


Dividing Polynomials using Long Division - Concept

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

When dividing polynomials, we can use either long division or synthetic division to arrive at an answer. Using long division, dividing polynomials is easy. We simply write the fraction in long division form by putting the divisor outside of the bracket and the divided inside the bracket. After the polynomial division is set up, we follow the same process as long division with numbers.

Dividing polynomials with long division. So we're going to talk about how we can divide these 2 polynomials using long division. Before we do that, I just want to profess it by you know drawing a parallel to something we already know how to do. Just say we're asked to divide 1,170 by 70, okay? What we end up doing is we draw our 7 in a little bracket with a number on the inside. We then start and say, okay. How many times does 7 go into 1, it doesn't so we go to the next [IB] How many times does 7 go into 11? Draw the 1, 1 times 7 is 7 and then we subtract dropping down the next term 47, continue from there. How many times does 7 go into 47, it's 6, 6 times 7 is 42,. Subtract again, leaving us 50 [IB] and repeat, we need 7, so this is in 49 and we're left with a remainder of 1. Okay.
So we know how to divide numbers. Okay? Dividing polynomials is exactly the same idea just with a little bit more x's and variables and terms involved. So what we're going to end up doing is exactly the same thing as this, okay? So x+1 goes on the outside and then our, remember this is called the dividend our dividend goes inside the bracket. Have a big bracket. 3x cubed and I think I forgot the square of my problem, let's throw that in there. 3x squared minus sorry 3x cubed minus 4x squared plus 2x minus 1, okay. So we basically are rewriting our problem as a fraction in long division. In logic it's exactly the same as we did right here, okay?
You say the first thing you want to get rid of is the 3x squared and we looked at our leading term on our divisor, the term in the bottom, okay? In order to get 3x squared from x we need to multiply it, sorry 3x cubed from x we need to multiply it by 3x squared. So I always try to line up my powers. So I'm going to put 3x squared right there. So I have all my squares in a column right here. Okay? Then like what we did over here we multiply the number on top by the number in front. The only difference now is we have some variables the same idea holds. Okay? So 3x squared times x, is 3x cubed. 3x squared times 1 is 3x squared. Okay? Just like we did over here with our numbers, we need to subtract, okay? So subtraction. 3x cubed minus 3x cubed, those cancel which is what we wanted and the -4x squared minus 3x, make sure you distribute that negative sign becomes negative 7x squared.
The next term we want to get rid of is -7x squared. If you want to bring this down, you can, you don't have to. Just remember that we need to include it next step when we subtract. Okay.
So we need to get rid of the -7x squared with our leading term of an x. So we need to multiply our x by -7x in order to cancel it out. -7x and then we just want to multiply and subtract once again. So -7x times x, -7x squared. -7x times 1 is -7x. Once again we want to subtract making sure we distribute that negative sign through. -7x squared minus -7x squared, those cancel which is what we wanted and then we still have these two up here. So it's 2x minus -7x, 2x plus 7x which turns into 9x and again we can bring this one down if we want, we don't need to. Just remember it's there. So this 9x is the last thing we need to get rid of. In order to get 9x from an x we need a 9 plus 9, that 9 gets distributed through. 9x plus 9 and once again we subtract, okay? 9x-9x. Those cancel. -1-9 plus a negative 9 so this is going to give us -10, which has a remainder [IB] we multiply x by will give us -10. Okay?
So there's different ways you can write this up. I'm going to take a second and come back over here and show us what this actually means. So, this was our process for numbers same exact process over there. Well, what we just really found out was that this statement is actually equal to our quotient. What comes up is the number that is above the sign, so if we come back over here. It's just a number up here. So that is 3x squared minus 7x plus 9 and then plus our remainder over our divisor. So this is then going to be plus -10 over our divisor here, x+1, okay? So long division is a little bit more in dealing with numbers but the process is exactly the same.


Mathematics Part I Solutions for Class 9 Maths Chapter 3 - Polynomials

Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 3 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among Class 9 students for Maths Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 39:

Question 1:

State whether the given algebraic expressions are polynomials? Justify.

(i) y   +   1 y (ii) 2   -   5   x (iii) x 2   +   7 x   +   9 (iv) 2 m - 2   +   7 m   -   5 (v) 10

Answer:


In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

Here, one of the powers of y is &minus1, which is not a whole number. So, y   +   1 y is not a polynomial.

Here, the power of x is 1 2 , which is not a whole number. So, 2   -   5   x is not a polynomial.

Here, the powers of the variable x are 2, 1 and 0, which are whole numbers. So, x 2   +   7 x   +   9 is a polynomial.

Here, one of the powers of m is &minus2, which is not a whole number. So, 2 m - 2   +   7 m   -   5 is not a polynomial.

Here, the power of x is 0, which is a whole numbers. So, 10 is a polynomial (or constant polynomial).

Page No 39:

Question 2:

Write the coefficient of m 3 in each of the given polynomial.

(i) m 3 (ii) - 3   2   +   m   -   3 m 3 (iii) - 2 3 m 3   -   5 m 2   +   7 m   -   1

Answer:

Page No 39:

Question 3:

Answer:


(i) A polynomial having only one term is called a monomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

3x 7 is a monomial in x with degree 7.

(ii) A polynomial having only two terms is called a binomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

2x 35 + 1 is a binomial in x with degree 35.

(iii) A polynomial having only three terms is called a trinomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.

5x 8 + 6x 4 + 7x is a trinomial in x with degree 8.

Page No 40:

Question 4:

Write the degree of the given polynomials.

(i) 5 (ii) x ° (iii) x 2 (iv) 2   m 10   -   7 (v) 2 p   -   7 (vi) 7 y   -   y 3   +   y 5 (vii) x y z   + x y   -   z (viii) m 3 n 7   -   3 m 5 n   +   m n

Answer:


The highest power of the variable in a polynomial of one variable is called the degreee of the polynomial. Also, the highest sum of the powers of the variables in each term of the polynomial in more than one variable is the degree of the polynomial.

The degree of the polynomial 5 is 0.

(ii)
The degree of the polynomial x 0 is 0.

(iii)
The degree of the polynomial x 2 is 2.

(iv)
The degree of the polynomial 2 m 10 - 7 is 10.

(v)
The degree of the polynomial 2 p - 7 is 1.

(vi)
The degree of the polynomial 7 y - y 3 + y 5 is 5.

(vii)
The sum of the powers of the variables in the polynomial x y z + x y - z are 1 + 1 + 1 = 3 and 1 + 1 = 2.

The degree of the polynomial x y z + x y - z is 3.

(viii)
The sum of the powers of the variables in the polynomial m 3 n 7 - 3 m 5 n + m n are 3 + 7 = 10, 5 + 1 = 6 and 1 + 1 = 2.

The degree of the polynomial m 3 n 7 - 3 m 5 n + m n is 10.

Page No 40:

Question 5:

Classify the following polynomials as linear, quadratic and cubic polynomial.

Answer:


(i)
The degree of the polynomial 2x 2 + 3x + 1 is 2.

So, the polynomial 2x 2 + 3x + 1 is a quadratic polynomial.

(ii)
The degree of the polynomial 5p is 1.

So, the polynomial 5p is a linear polynomial.

(iii)
The degree of the polynomial 2 y - 1 2 is 1.

So, the polynomial 2 y - 1 2 is a linear polynomial.

(iv)
The degree of the polynomial m 3 + 7 m 2 + 5 2 m - 7 is 3.

So, the polynomial m 3 + 7 m 2 + 5 2 m - 7 is a cubic polynomial.

(v)
The degree of the polynomial a 2 is 2.

So, the polynomial a 2 is a quadratic polynomial.

(vi)
The degree of the polynomial 3r 3 is 3.

So, the polynomial 3r 3 is a cubic polynomial.

Page No 40:

Question 6:

Write the following polynomials in standard form.

(i) m 3   +   3   +   5 m (ii) - 7 y   +   y 5   +   3 y 3   -   1 2   +   2 y 4   -   y 2

Answer:


A polynomial written in either descending or ascending powers of its variable is called the standard form of the polynomial.

(i)
The given polynomial is m 3 + 3 + 5m.

The standard form of the polynomial is 3 + 5m + m 3 or m 3 + 5m + 3.

(ii)
The given polynomial is - 7 y + y 5 + 3 y 3 - 1 2 + 2 y 4 - y 2 .

The standard form of the polynomial is y 5 + 2 y 4 + 3 y 3 - y 2 - 7 y - 1 2 or - 1 2 - 7 y - y 2 + 3 y 3 + 2 y 4 + y 5 .

Page No 40:

Question 7:

Write the following polynomials in coefficient form.

(i) x 3   -   2 (ii) 5 y (iii) 2 m 4   -   3 m 2   +   7 (iv)   - 2 3

Answer:


(i)
x 3 - 2 = x 3 + 0 x 2 + 0 x - 2

The coefficient form of the polynomial is (1, 0, 0, &minus2).

The coefficient form of the polynomial is (5, 0).

(iii)
2 m 4 - 3 m 2 + 7 = 2 m 4 + 0 m 3 - 3 m 2 + 0 m + 7

The coefficient form of the polynomial is (2, 0, &minus3, 0, 7).

(iv)
The coefficient form of the polynomial - 2 3 is - 2 3 .

Page No 40:

Question 8:

Answer:


(i)
The coefficient form of the polynomial is (1, 2, 3).

Therefore, the index form of the polynomial is x 2 + 2x + 3.

(ii)
The coefficient form of the polynomial is (5, 0, 0, 0, &minus1).

Therefore, the index form of the polynomial is 5x 4 + 0x 3 + 0x 2 + 0x &minus 1 or 5x 4 &minus 1.

(iii)
The coefficient form of the polynomial is (&minus2, 2, &minus2, 2).

Therefore, the index form of the polynomial is &minus2x 3 + 2x 2 &minus2x + 2.

Page No 40:

Question 9:

Write the appropriate polynomials in the boxes.

Answer:

Page No 43:

Question 1:

Answer:


(i)
Initial number of trees in the village = a

Increase in the number of trees every year = b

&there4 Number of trees in the village after x years

= Initial number of trees in the village + Increase in the number of trees every year × x

Thus, the number of trees after x years is a + bx.

(ii)
Number of students in each row = y

&there4 Total number of students in the parade = Number of students in each row × Number of rows = y × x = yx = xy

Thus, there are in all xy students for the parade.

(iii)
Digit at the tens place = m

Digit at the units place = n

&there4 Two digit number = Digit at the tens place × 10 + Digit at the units place = m × 10 + n = 10m + n

Thus, the polynomial which represents the two digit number is 10m + n.

Page No 43:

Question 2:

Add the given polynomials.

Answer:


(i)
x 3 - 2 x 2 - 9 + 5 x 3 + 2 x + 9 = x 3 + 5 x 3 - 2 x 2 + 2 x - 9 + 9 = 6 x 3 - 2 x 2 + 2 x
(ii)
- 7 m 4 + 5 m 3 + 2 + 5 m 4 - 3 m 3 + 2 m 2 + 3 m - 6 = - 7 m 4 + 5 m 4 + 5 m 3 - 3 m 3 + 2 m 2 + 3 m + 2 - 6 = - 2 m 4 + 2 m 3 + 2 m 2 + 3 m - 6 + 2
(iii)
2 y 2 + 7 y + 5 + 3 y + 9 + 3 y 2 - 4 y - 3 = 2 y 2 + 3 y 2 + 7 y + 3 y - 4 y + 5 + 9 - 3 = 5 y 2 + 6 y + 11

Page No 43:

Question 3:

Subtract the second polynomial from the first.

(i) x 2   -   9 x   +   3       - 19 x   +   3   +   7 x 2 (ii) 2 a b 2   +   3 a 2 b   -   4 a b       3 a b   - 8 a b 2   +   2 a 2 b

Answer:


(i)
x 2 - 9 x + 3   - - 19 x + 3 + 7 x 2 = x 2 - 9 x + 3 + 19 x - 3 - 7 x 2 = x 2 - 7 x 2 - 9 x + 19 x + 3 - 3 = - 6 x 2 + 10 x
(ii)
2 a b 2 + 3 a 2 b - 4 a b - 3 a b - 8 a b 2 + 2 a 2 b = 2 a b 2 + 3 a 2 b - 4 a b - 3 a b + 8 a b 2 - 2 a 2 b = 2 a b 2 + 8 a b 2 + 3 a 2 b - 2 a 2 b - 4 a b - 3 a b = 10 a b 2 + a 2 b - 7 a b

Page No 43:

Question 4:

Multiply the given polynomials.

Answer:


(i)
2 x x 2   - 2 x - 1 = 2 x × x 2 + 2 x × - 2 x + 2 x × - 1 = 2 x 3 - 4 x 2 - 2 x
(ii)
x 5 - 1 × x 3 + 2 x 2 + 2 = x 5 x 3 + 2 x 2 + 2 - 1 x 3 + 2 x 2 + 2 = x 8 + 2 x 7 + 2 x 5 - x 3 - 2 x 2 - 2
(iii)
2 y + 1 × y 2 - 2 y 3 + 3 y = 2 y y 2 - 2 y 3 + 3 y + 1 y 2 - 2 y 3 + 3 y = 2 y 3 - 4 y 4 + 6 y 2 + y 2 - 2 y 3 + 3 y = - 4 y 4 + 2 y 3 - 2 y 3 + 6 y 2 + y 2 + 3 y = - 4 y 4 + 7 y 2 + 3 y

Page No 43:

Question 5:

Answer:

(i)
x 3 - 64 = x 3 + 0 x 2 + 0 x - 64

Using long division method,


Dividend = Divisor ×​ Quotient + Remainder

∴ x 3 - 64 = x - 4 × x 2 + 4 x + 16 + 0

(ii)
5 x 5 + 4 x 4 - 3 x 3 + 2 x 2   + 2 = 5 x 5 + 4 x 4 - 3 x 3 + 2 x 2   + 0 x + 2

Using long division method,


Dividend = Divisor ×​ Quotient + Remainder

∴ 5 x 5 + 4 x 4 - 3 x 3 + 2 x 2 + 2 = x 2 - x × 5 x 3 + 9 x 2 + 6 x + 8 + 8 x + 2

Page No 43:

Question 6:

Answer:


Lenght of the rectangular farm = (2a 2 + 3b 2 ) m

Breadth of the rectangular farm = (a 2 + b 2 ) m

Side of the square plot = (a 2 &minus b 2 ) m

&there4 Area of the remaining part of the farm

= Total area of the farm &minus Area of the square plot

= Lenght of the rectangular farm × Breadth of the rectangular farm &minus (Side of the square plot) 2

Thus, the area of the remaining part of the farm is (a 4 + 7a 2 b 2 + 2b 4 ) m 2 .

Page No 46:

Question 1:

linear division method. Write the quotient and the remainder.

Answer:

(i)
Synthetic Division:


The coefficient form of the quotient is (2, 7).

&there4 Quotient = 2m + 7 and Remainder = 45

Linear Method:

2 m 2 - 3 m + 10 = 2 m m - 5 + 10 m - 3 m + 10 = 2 m m - 5 + 7 m - 5 + 35 + 10 = m - 5 × 2 m + 7 + 45
(ii)
Synthetic Division:

Dividend = x 4 + 2 x 3 + 3 x 2 + 4 x + 5

The coefficient form of the quotient is (1, 0, 3, &minus2).

&there4 Quotient = x 3 + 3x &minus 2 and Remainder = 9

Linear Method:

x 4 + 2 x 3 + 3 x 2 + 4 x + 5 = x 3 x + 2 + 3 x x + 2 - 6 x + 4 x + 5 = x 3 x + 2 + 3 x x + 2 - 2 x + 5 = x 3 x + 2 + 3 x x + 2 - 2 x + 2 + 4 + 5 = x + 2 × x 3 + 3 x - 2 + 9
(iii)
Synthetic Division:

Dividend = y 3 - 216 = y 3 + 0 y 2 + 0 y - 216

The coefficient form of the quotient is (1, 6, 36).

&there4 Quotient = y 2 + 6y + 36 and Remainder = 0

Linear Method:

y 3 - 216 = y 2 y - 6 + 6 y 2 - 216 = y 2 y - 6 + 6 y y - 6 + 36 y - 216 = y 2 y - 6 + 6 y y - 6 + 36 y - 6 + 216 - 216 = y 2 y - 6 + 6 y y - 6 + 36 y - 6 = y - 6 × y 2 + 6 y + 36
(iv)
Synthetic Division:

Dividend = 2 x 4 + 3 x 3 + 4 x - 2 x 2 = 2 x 4 + 3 x 3 - 2 x 2 + 4 x + 0

The coefficient form of the quotient is (2, &minus3, 7, &minus17).

&there4 Quotient = 2x 3 &minus 3x 2 + 7x &minus 17 and Remainder = 51

Linear Method:

2 x 4 + 3 x 3 - 2 x 2 + 4 x = 2 x 3 x + 3 - 6 x 3 + 3 x 3 - 2 x 2 + 4 x = 2 x 3 x + 3 - 3 x 2 x + 3 + 9 x 2 - 2 x 2 + 4 x = 2 x 3 x + 3 - 3 x 2 x + 3 + 7 x x + 3 - 21 x + 4 x = 2 x 3 x + 3 - 3 x 2 x + 3 + 7 x x + 3 - 17 x + 3 + 51 = x + 3 × 2 x 3 - 3 x 2 + 7 x - 17 + 51
(v)
Synthetic Division:

Dividend = x 4 - 3 x 2 - 8 = x 4 + 0 x 3 - 3 x 2 + 0 x - 8

The coefficient form of the quotient is (1, &minus4, 13, &minus52).

&there4 Quotient = x 3 &minus 4x 2 + 13x &minus 52 and Remainder = 200

Linear Method:

x 4 - 3 x 2 - 8 = x 3 x + 4 - 4 x 3 - 3 x 2 - 8 = x 3 x + 4 - 4 x 2 x + 4 + 16 x 2 - 3 x 2 - 8 = x 3 x + 4 - 4 x 2 x + 4 + 13 x x + 4 - 52 x - 8 = x 3 x + 4 - 4 x 2 x + 4 + 13 x x + 4 - 52 x + 4 + 208 - 8 = x + 4 × x 3 - 4 x 2 + 13 x - 52 + 200
(vi)
Synthetic Division:

Dividend = y 3 - 3 y 2 + 5 y - 1

The coefficient form of the quotient is (1, &minus2, 3).

&there4 Quotient = y 2 &minus 2y + 3 and Remainder = 2

Linear Method:

y 3 - 3 y 2 + 5 y - 1 = y 2 y - 1 + y 2 - 3 y 2 + 5 y - 1 = y 2 y - 1 - 2 y y - 1 - 2 y + 5 y - 1 = y 2 y - 1 - 2 y y - 1 + 3 y - 1 + 3 - 1 = y - 1 × y 2 - 2 y + 3 + 2

Page No 48:

Question 1:

For x = 0 find the value of the polynomial x 2   -   5 x   +   5 .

Answer:

∴ p 0 = 0 2 - 5 × 0 + 5 = 0 - 0 + 5 = 5

Hence, for x = 0 the value of the polynomial is 5.

Page No 48:

Question 2:

If p y   =   y 2   -   3 2 y   +   1   then find p   3 2 .

Answer:

∴ p 3 2 = 3 2 2 - 3 2 × 3 2 + 1 = 18 - 18 + 1 = 1

Page No 48:

Question 3:

Answer:

∴ p a = a 3 + 2 a 2 - a + 10 . (1)

p - a = - a 3 + 2 - a 2 - - a + 10

⇒ p - a = - a 3 + 2 a 2 + a + 10 . (2)

p a + p - a = a 3 + 2 a 2 - a + 10 + - a 3 + 2 a 2 + a + 10 = a 3 - a 3 + 2 a 2 + 2 a 2 - a + a + 10 + 10 = 4 a 2 + 20
∴ p a + p - a = 4 a 2 + 20

Page No 48:

Question 4:

If p y   =   2 y 3   -   6 y 2   -   5 y   +   7 then find p 2 .

Answer:


p y = 2 y 3 - 6 y 2 - 5 y + 7

∴ p 2 = 2 × 2 3 - 6 × 2 2 - 5 × 2 + 7 = 16 - 24 - 10 + 7 = - 11

Page No 53:

Question 1:

Find the value of the polynomial 2 x   -   2 x 3   +   7 using given values for x .

Answer:

(i)
p 3 = 2 × 3 - 2 × 3 3 + 7 = 6 - 2 × 27 + 7 = 6 - 54 + 7 = - 41
Thus, the value of polynomial for x = 3 is &minus41.

(ii)
p - 1 = 2 × - 1 - 2 × - 1 3 + 7 = - 2 - 2 × - 1 + 7 = - 2 + 2 + 7 = 7
Thus, the value of polynomial for x = &minus1 is 7.

(iii)
p 0 = 2 × 0 - 2 × 0 3 + 7 = 0 - 0 + 7 = 7
Thus, the value of polynomial for x = 0 is 7.

Page No 53:

Question 2:

For each of the following polynomial, find p 1 ,   p 0   and p - 2 .

(i) p x   =   x 3 (ii) p y   =   y 2   - 2 y   +   5   (iii) p x   =   x 4   - 2 x 2   -   x

Answer:


(i)
p x = x 3 ∴ p 1 = 1 3 = 1 p 0 = 0 3 = 0 p - 2 = - 2 3 = - 8
(ii)
p y = y 2 - 2 y + 5 ∴ p 1 = 1 2 - 2 × 1 + 5 = 1 - 2 + 5 = 4 p 0 = 0 2 - 2 × 0 + 5 = 0 - 0 + 5 = 5 p - 2 = - 2 2 - 2 × - 2 + 5 = 4 + 4 + 5 = 13
(iii)
p x = x 4 - 2 x 2 - x ∴ p 1 = 1 4 - 2 × 1 2 - 1 = 1 - 2 - 1 = - 2 p 0 = 0 4 - 2 × 0 2 - 0 = 0 - 0 - 0 = 0 p - 2 = - 2 4 - 2 × - 2 2 - - 2 = 16 - 2 × 4 + 2 = 16 - 8 + 2 = 10

Page No 53:

Question 3:

If the value of the polynomial m 3   +   2 m   +   a is 12 for m   =   2 , then find the value of a.

Answer:

∴ 2 3 + 2 × 2 + a = 12 ⇒ 8 + 4 + a = 12 ⇒ 12 + a = 12 ⇒ a = 12 - 12 = 0
Thus, the value of a is 0.

Page No 53:

Question 4:

For the polynomial m x 2   - 2 x   +   3 if p - 1   = 7   then find m.

Answer:

∴ p - 1 = 7 ⇒ m × - 1 2 - 2 × - 1 + 3 = 7 ⇒ m + 2 + 3 = 7 ⇒ m = 7 - 5 = 2
Thus, the value of m is 2.

Page No 53:

Question 5:

Divide the first polynomial by the second polynomial and find the remainder using factor theorem .

Answer:

(i)
By synthetic division:

The coefficient form of the quotient is (1, &minus8).

By remainder theorem:

Remainder = p(&minus1) = - 1 2 - 7 × - 1 + 9 = 1 + 7 + 9 = 19

(ii)
By synthetic division:

Dividend = 2 x 3 - 2 x 2 + a x - a

The coefficient form of the quotient is (2, 2a &minus 2, 2a 2 ​ &minus a).

By remainder theorem:

Let p x = 2 x 3 - 2 x 2 + a x - a .

Remainder = p(a) = 2 × a 3 - 2 × a 2 + a × a - a = 2 a 3 - 2 a 2 + a 2 - a = 2 a 3 - a 2 - a

(iii)
By synthetic division:

Dividend = 54 m 3 + 18 m 2 - 27 m + 5

The coefficient form of the quotient is (54, 180, 513).

By remainder theorem:

Let p m = 54 m 3 + 18 m 2 - 27 m + 5 .

Remainder = p(3) = 54 × 3 3 + 18 × 3 2 - 27 × 3 + 5 = 54 × 27 + 18 × 9 - 27 × 3 + 5 = 1458 + 162 - 81 + 5 = 1544

Page No 53:

Question 6:

If the polynomial y 3   -   5 y 2   +   7 y   +   m is divided by y + 2 and the remainder is 50 then find the value of m.

Answer:


Let p y = y 3 - 5 y 2 + 7 y + m .

When the polynomial is divided by (y + 2), the remainder is 50. This means that the value of the polynomial when y = &minus2 is 50.

∴ - 2 3 - 5 × - 2 2 + 7 × - 2 + m = 50 ⇒ - 8 - 5 × 4 - 14 + m = 50 ⇒ - 8 - 20 - 14 + m = 50 ⇒ - 42 + m = 50 ⇒ m = 50 + 42 = 92
Thus, the value of m is 92.

Page No 53:

Question 7:

Use factor theorem to determine whether x + 3 is factor of x 2 + 2x &minus 3 or not.

Answer:

∴ p - 3 = - 3 2 + 2 × - 3 - 3 = 9 - 6 - 3 = 0

So, by factor theorem, (x + 3) is a factor of x 2 + 2x &minus 3.

Page No 53:

Question 8:

If ( x -   2 ) is a factor of x 3   -   m x 2   +   10 x   -   20   then find the value of m.

Answer:


Let p x = x 3 - m x 2 + 10 x - 20 .

It is given that (x &minus 2) is a factor of p x = x 3 - m x 2 + 10 x - 20 .

∴ p 2 = 0 ⇒ 2 3 - m × 2 2 + 10 × 2 - 20 = 0 ⇒ 8 - 4 m + 20 - 20 = 0 ⇒ 8 - 4 m = 0 ⇒ 4 m = 8 ⇒ m = 2
Thus, the value of m is 2.

Page No 53:

Question 9:

By using factor theorem in the following examples, determine whether q ( x ) is a factor p ( x ) or not.

Answer:

∴ p 1 = 1 3 - 1 2 - 1 - 1 = 1 - 1 - 1 - 1 = - 2

Since p(1) &ne 0, so by factor theorem q x = x - 1 is not a factor of polynomial p x = x 3 - x 2 - x - 1 .

∴ p 3 = 2 × 3 3 - 3 2 - 45 = 2 × 27 - 9 - 45 = 54 - 54 = 0

So, by factor theorem q x = x - 3 is a factor of polynomial p x = 2 x 3 - x 2 - 45 .

Page No 53:

Question 10:

If ( x 31 + 31) is divided by (x + 1) then find the remainder.

Answer:

By remainder theorem, we have

Remainder = p(&minus1) = (&minus1) 31 + 31 = &minus1 + 31 = 30

Thus, the remainder when (x 31 + 31) is divided by (x + 1) is 30.

Page No 53:

Question 11:

Show that m - 1 is a factor of m 21 - 1 and m 22 - 1.

Answer:

Therefore, by factor theorem (m &minus 1) is a factor of p(m) = m 21 &minus 1.

Therefore, by factor theorem (m &minus 1) is a factor of q(m) = m 22 &minus 1.

Hence, (m &minus 1) is a factor of m 21 &minus 1 and m 22 &minus 1.

Page No 53:

Question 12:

If   x   -   2   and x   -   1 2 both are the factors of the polynomial nx 2 &minus 5x + m, then show that m   =   n =   2

Answer:

It given that (x &minus 2) and x - 1 2 are the factors of the polynomial p(x) = nx 2 &minus 5x + m.

&there4 By factor theorem, p(2) = 0 and p 1 2 = 0 .

p 1 2 = 0 ⇒ n 1 2 2 - 5 × 1 2 + m = 0 ⇒ n 4 + m = 5 2 ⇒ n + 4 m = 10                                 . . . . . 2
From (1) and (2), we have

Putting n = m in (1), we have

Page No 53:

Question 13:

(i) If p x   =   2   +   5 x   then p 2   +   p - 2 -   p 1 .

(ii) p x   =   2 x 2   -   5 3 x   +   5   then p 5 3 .

Answer:


(i)
p x = 2 + 5 x ∴ p 2 = 2 + 5 × 2 = 2 + 10 = 12 p - 2 = 2 + 5 × - 2 = 2 - 10 = - 8 p 1 = 2 + 5 × 1 = 2 + 5 = 7
∴ p 2 + p - 2 - p 1 = 12 + - 8 - 7 = 12 - 8 - 7 = 12 - 15 = - 3

(ii)
p x = 2 x 2 - 5 3 x + 5
∴ p 5 3 = 2 × 5 3 2 - 5 3 × 5 3 + 5 = 2 × 75 - 25 × 3 + 5 = 150 - 75 + 5 = 80

Page No 54:

Question 1:

Find the factors of the polynomials given below.
(i) 2x 2 + x &ndash 1

Answer:


(i)
2 x 2 + x - 1 = 2 x 2 + 2 x - x - 1 = 2 x x + 1 - 1 x + 1 = x + 1 2 x - 1
(ii)
2 m 2 + 5 m - 3 = 2 m 2 + 6 m - m - 3 = 2 m m + 3 - 1 m + 3 = m + 3 2 m - 1
(iii)
12 x 2 + 61 x + 77 = 12 x 2 + 28 x + 33 x + 77 = 4 x 3 x + 7 + 11 3 x + 7 = 3 x + 7 4 x + 11
(iv)
3 y 2 - 2 y - 1 = 3 y 2 - 3 y + y - 1 = 3 y y - 1 + 1 y - 1 = y - 1 3 y + 1
(v)
3 x 2 + 4 x + 3 = 3 x 2 + 3 x + x + 3 = 3 x x + 3 + 1 x + 3 = x + 3 3 x + 1
(vi)
1 2 x 2 - 3 x + 4 = 1 2 x 2 - 2 x - x + 4 = 1 2 x x - 4 - 1 x - 4 = x - 4 1 2 x - 1

Page No 55:

Question 2:

Answer:


(i)
(x 2 &ndash x) 2 &ndash 8(x 2 &ndash x) + 12
Let x 2 &ndash x = z.
∴ x 2 - x 2 - 8 x 2 - x + 12 = z 2 - 8 z + 12 = z 2 - 6 z - 2 z + 12 = z z - 6 - 2 z - 6 = z - 6 z - 2
= x 2 - x - 6 x 2 - x - 2                                             Replace   z = x 2 - x = x 2 - 3 x + 2 x - 6 x 2 - 2 x + x - 2 = x x - 3 + 2 x - 3 x x - 2 + 1 x - 2 = x - 3 x + 2 x - 2 x + 1
(ii)
(x &ndash 5) 2 &ndash (5x &ndash 25) &ndash 24
= (x &ndash 5) 2 &ndash 5(x &ndash 5) &ndash 24
Let x &ndash 5 = z.
∴ x - 5 2 - 5 x - 5 - 24 = z 2 - 5 z - 24 = z 2 - 8 z + 3 z - 24 = z z - 8 + 3 z - 8 = z - 8 z + 3
= x - 5 - 8 x - 5 + 3                                   Replace   z = x - 5 = x - 13 x - 2
(iii)
(x 2 &ndash 6x) 2 &ndash 8(x 2 &ndash 6x + 8) &ndash 64
= (x 2 &ndash 6x) 2 &ndash 8(x 2 &ndash 6x) &ndash 64 &ndash 64
= (x 2 &ndash 6x) 2 &ndash 8(x 2 &ndash 6x) &ndash 128
Let x 2 &ndash 6x = z.
∴ x 2 - 6 x 2 - 8 x 2 - 6 x - 128 = z 2 - 8 z - 128 = z 2 - 16 z + 8 z - 128 = z z - 16 + 8 z - 16 = z - 16 z + 8
= x 2 - 6 x - 16 x 2 - 6 x + 8                                   Replace   z = x 2 - 6 x = x 2 - 8 x + 2 x - 16 x 2 - 4 x - 2 x + 8 = x x - 8 + 2 x - 8 x x - 4 - 2 x - 4 = x - 8 x + 2 x - 4 x - 2

(iv)
(x 2 &ndash 2x + 3)(x 2 &ndash 2x + 5) &ndash 35
Let x 2 &ndash 2x = z.
∴ x 2 - 2 x + 3 x 2 - 2 x + 5 - 35 = z + 3 z + 5 - 35 = z 2 + 5 z + 3 z + 15 - 35 = z 2 + 8 z - 20 = z 2 + 10 z - 2 z - 20
= z z + 10 - 2 z + 10 = z + 10 z - 2 = x 2 - 2 x + 10 x 2 - 2 x - 2                               Replace   z = x 2 - 2 x
(v)
(y + 2)(y &ndash 3)(y + 8)(y + 3) + 56
= (y + 2)(y + 3)(y + 8)(y &ndash 3) + 56
= (y 2 + 5y + 6)(y 2 + 5y &ndash 24) + 56
Let y 2 + 5y = z.
∴ y 2 + 5 y + 6 y 2 + 5 y - 24 + 56 = z + 6 z - 24 + 56 = z 2 - 18 z - 144 + 56 = z 2 - 18 z - 88
= z 2 - 22 z + 4 z - 88 = z z - 22 + 4 z - 22 = z - 22 z + 4 = y 2 + 5 y - 22 y 2 + 5 y + 4                           Replace   z = y 2 + 5 y
= y 2 + 5 y - 22 y 2 + 4 y + y + 4 = y 2 + 5 y - 22 y y + 4 + 1 y + 4 = y 2 + 5 y - 22 y + 4 y + 1
(vi)
(y 2 + 5y)(y 2 + 5y &ndash 2) &ndash 24
Let y 2 + 5y = z.
∴ y 2 + 5 y y 2 + 5 y - 2 - 24 = z z - 2 - 24 = z 2 - 2 z - 24 = z 2 - 6 z + 4 z - 24 = z z - 6 + 4 z - 6 = z - 6 z + 4
= y 2 + 5 y - 6 y 2 + 5 y + 4                                   Replace   z = y 2 + 5 y = y 2 + 6 y - y - 6 y 2 + 4 y + y + 4 = y y + 6 - 1 y + 6 y y + 4 + 1 y + 4 = y + 6 y - 1 y + 4 y + 1
(vii)
(x &ndash 3)(x &ndash 4) 2 (x &ndash 5) &ndash 6
= (x &ndash 3)(x &ndash 5)(x &ndash 4) 2 &ndash 6
= (x 2 &ndash 8x + 15)(x 2 &ndash 8x + 16) &ndash 6
Let x 2 &ndash 8x = z.
∴ x 2 - 8 x + 15 x 2 - 8 x + 16 - 6 = z + 15 z + 16 - 6 = z 2 + 31 z + 240 - 6 = z 2 + 31 z + 234
= z 2 + 18 z + 13 z + 234 = z z + 18 + 13 z + 18 = z + 18 z + 13 = x 2 - 8 x + 18 x 2 - 8 x + 13                                   Replace   z = x 2 - 8 x

Page No 55:

Question 1:

(i) Which of the following is a polynomial ?
(A) x y (B) x 2   -   3 x (C) x - 2   +   7 (D) 2 x 2   +   1 2

(ii) What is the degree of the polynomial 7 ?
(A) 1 2 (B) 5 (C) 2 (D) 0

(iii) What is the degree of the 0 polynomial ?

(A) 0 (B) 1 (C) undefined (D) any real number

(iv) What is the degree of the polynomial 2x 2 + 5x 3 + 7?
(A) 3 (B) 2 (C) 5 (D) 7

(v) What is the coefficient form of   x 3   -   1 ?
(A) (1, - 1) (B) (3, - 1) (C) (1, 0, 0, - 1) (D) (1, 3, - 1)

(vi) p ( x )   =   x 2   -   7 7 x   +   3 then p 7 7   =   ?
(A) 3 (B) 7 7 (C) 42   7   +   3 (D) 49 7

(vii) When x   =   -   1 , what is the value of the polynomial 2x 3 + 2x ?
(A) 4 (B) 2 (C) - 2 (D) - 4

(viii) If x - 1   , what is a factor of the polynomial 3 x 2   +   m x then find the value of m.
(A) 2 (B) - 2 (C) - 3 (D) 3

(ix) Multiply ( x 2 - 3) (2x - 7x 3 + 4) and write the degree of the product.
(A) 5 (B) 3 (C) 2 (D) 0

(x) Which of the following is a linear polynomial ?

Answer:


(i)
In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.

In the expression 2 x 2   +   1 2 , the power of the variable x is 2, which is a whole number. So, the expression 2 x 2   +   1 2 is a polynomial.

Hence, the correct answer is option (D).

The degree of the polynomial 7 is 0.

Hence, the correct answer is option (D).

(iii)
The degree of 0 polynomial is not defined.

Hence, the correct answer is option (C).

(iv)
The highest power of the variable in a polynomial is called the degree of the polynomial.

The degree of the polynomial 2x 2 + 5x 3 + 7 is 3.

Hence, the correct answer is option (A).

(v)
  x 3 - 1 = x 3 + 0 x 2 + 0 x - 1

The coefficient form of the polynomial   x 3 - 1 is (1, 0, 0, &minus1).

Hence, the correct answer is option (C).

∴ p 7 7 =   7 7 2 - 7 7 × 7 7 + 3 = 343 - 343 + 3 = 3

Hence, the correct answer is option (A).

Thus, the value of the polynomial when x = &minus1 is &minus4.

Hence, the correct answer is option (D).

Hence, the correct answer is option (C).

(ix)
x 2 - 3 2 x - 7 x 3 + 4 = x 2 2 x - 7 x 3 + 4 - 3 2 x - 7 x 3 + 4 = 2 x 3 - 7 x 5 + 4 x 2 - 6 x + 21 x 3 - 12 = - 7 x 5 + 2 x 3 + 21 x 3 + 4 x 2 - 6 x - 12 = - 7 x 5 + 23 x 3 + 4 x 2 - 6 x - 12
Thus, the degree of the resultant polynomial is 5.

Hence, the correct answer is option (A).

(x)
A polynomial with degree one is called a linear polynomial.

Thus, the polynomial x + 5 is a linear polynomial.

Hence, the correct answer is option (A).

Page No 56:

Question 2:

Write the degree of the polynomial for each of the following.

Answer:


The highest power of the variable in a polynomial in one variable is called the degree of the polynomial.

(i)
The degree of the polynomial 5 + 3x 4 is 4.

The degree of the constant polynomial 7 is 0.

(iii)
The degree of the polynomial ax 7 + bx 9 is 9.

Page No 56:

Question 3:

Answer:


A polynomial written in either descending or ascending power of its variable is called the standard form of the polynomial.

(i)
The given polynomial is 4x 2 + 7x 4 &minus x 3 &minus x + 9 .
The standard form of the polynomial is 7x 4 &minus x 3 + 4x 2 &minus x + 9 or 9 &minus x + 4x 2 &minus x 3 + 7 x 4 .

(ii)
The given polynomial is p + 2p 3 + 10p 2 + 5p 4 &minus 8.

The standard form of the polynomial is 5p 4 + 2p 3 + 10p 2 + p &minus 8 or &minus8 + p + 10p 2 + 2p 3 + 5p 4 .

Page No 56:

Question 4:

Answer:


(i)
x 4 + 16 = x 4 + 0 x 3 + 0 x 2 + 0 x + 16

Therefore, the given polynomial in coefficient form is (1, 0, 0, 0, 16).

(ii)
m 5 + 2 m 2 + 3 m + 15 = m 5 + 0 m 4 + 0 m 3 + 2 m 2 + 3 m + 15

Therefore, the given polynomial in coefficient form is (1, 0, 0, 2, 3, 15).

Page No 56:

Question 5:

Answer:


(i)
The coefficient form of the polynomial is (3, &minus2, 0, 7, 18).

Therefore, the index form the polynomial is 3x 4 &minus 2x 3 + 0x 2 + 7x + 18 or 3x 4 &minus 2x 3 + 7x + 18.

(ii)
The coefficient form of the polynomial is (6, 1, 0, 7).

Therefore, the index form the polynomial is 6x 3 + x 2 + 0x + 7 or 6x 3 + x 2 + 7.

(iii)
The coefficient form of the polynomial is (4, 5, &minus3, 0).

Therefore, the index form the polynomial is 4x 3 + 5x 2 &minus 3x + 0 or 4x 3 + 5x 2 &minus 3x.

Page No 56:

Question 6:

Answer:


(i)
7 x 4 - 2 x 3 + x + 10 + 3 x 4 + 15 x 3 + 9 x 2 - 8 x + 2 = 7 x 4 + 3 x 4 - 2 x 3 + 15 x 3 + 9 x 2 + x - 8 x + 10 + 2 = 10 x 4 + 13 x 3 + 9 x 2 - 7 x + 12
(ii)
3 p 3 q + 2 p 2 q + 7 + 2 p 2 q + 4 p q - 2 p 3 q = 3 p 3 q - 2 p 3 q + 2 p 2 q + 2 p 2 q + 4 p q + 7 = p 3 q + 4 p 2 q + 4 p q + 7

Page No 56:

Question 7:

Answer:


(i)
5 x 2 - 2 y + 9 - 3 x 2 + 5 y - 7 = 5 x 2 - 2 y + 9 - 3 x 2 - 5 y + 7 = 5 x 2 - 3 x 2 - 2 y - 5 y + 9 + 7 = 2 x 2 - 7 y + 16
(ii)
2 x 2 + 3 x + 5 - x 2 - 2 x + 3 = 2 x 2 + 3 x + 5 - x 2 + 2 x - 3 = 2 x 2 - x 2 + 3 x + 2 x + 5 - 3 = x 2 + 5 x + 2

Page No 56:

Question 8:

Answer:


(i)
m 3 - 2 m + 3 m 4 - 2 m 2 + 3 m + 2 = m 3 m 4 - 2 m 2 + 3 m + 2 - 2 m m 4 - 2 m 2 + 3 m + 2 + 3 m 4 - 2 m 2 + 3 m + 2 = m 7 - 2 m 5 + 3 m 4 + 2 m 3 - 2 m 5 + 4 m 3 - 6 m 2 - 4 m + 3 m 4 - 6 m 2 + 9 m + 6 = m 7 - 2 m 5 - 2 m 5 + 3 m 4 + 3 m 4 + 2 m 3 + 4 m 3 - 6 m 2 - 6 m 2 - 4 m + 9 m + 6 = m 7 - 4 m 5 + 6 m 4 + 6 m 3 - 12 m 2 + 5 m + 6
(ii)
5 m 3 - 2 m 2 - m + 3 = 5 m 3 m 2 - m + 3 - 2 m 2 - m + 3 = 5 m 5 - 5 m 4 + 15 m 3 - 2 m 2 + 2 m - 6

Page No 56:

Question 9:

Answer:

Dividend = 3 x 3 - 8 x 2 + x + 7


The coefficient form of the quotient is (3, 1, 4).

&there4 Quotient = 3x 2 + x + 4 and Remainder = 19

Page No 56:

Question 10:

For which the value of m, x + 3 is the factor of the polynomial x 3 - 2mx + 21?

Answer:

(x + 3) is the factor of the polynomial p x = x 3 - 2 m x + 21 .

∴ p - 3 = 0 ⇒ - 3 3 - 2 m × - 3 + 21 = 0 ⇒ - 27 + 6 m + 21 = 0 ⇒ 6 m - 6 = 0 ⇒ 6 m = 6 ⇒ m = 1
Thus, the value of m is 1.

Page No 56:

Question 11:

At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x 2 - 3 y 2 , 7 y 2 + 2 xy and 9 x 2 + 4 xy respectively. At the beginning of the year 2017, x 2 + xy - y 2 , 5 xy and 3 x 2 + xy persons from each of the three villages respectively went to another village for education then what is the remaining total population of these three villages ?

Answer:


Total population of the three villages
= 5 x 2 - 3 y 2 + 7 y 2 + 2 x y + 9 x 2 + 4 x y = 5 x 2 + 9 x 2 - 3 y 2 + 7 y 2 + 2 x y + 4 x y = 14 x 2 + 4 y 2 + 6 x y
Total number of persons who went to another village for education
= x 2 + x y - y 2 + 5 x y + 3 x 2 + x y = x 2 + 3 x 2 - y 2 + x y + 5 x y + x y = 4 x 2 - y 2 + 7 x y
&there4 Remaining total population of the three villages = Total population of the three villages &minus Total number of persons who went to another village for education
= 14 x 2 + 4 y 2 + 6 x y - 4 x 2 - y 2 + 7 x y = 14 x 2 + 4 y 2 + 6 x y - 4 x 2 + y 2 - 7 x y = 14 x 2 - 4 x 2 + 4 y 2 + y 2 + 6 x y - 7 x y = 10 x 2 + 5 y 2 - x y
Thus, the remaining total population of these three villages is 10x 2 + 5y 2 &minus xy.

Page No 56:

Question 12:

Polynomials bx 2 + x + 5 and bx 3 - 2x + 5 are divided by polynomial x - 3 and the remainders are m and n respectively. If m - n = 0 then find the value of b.

Answer:


Let p x = b x 2 + x + 5 and q x = b x 3 - 2 x + 5 .

The remainder when p x = b x 2 + x + 5 is divided by (x &minus 3) is m.

∴ b × 3 3 - 2 × 3 + 5 = n ⇒ n = 27 b - 6 + 5 ⇒ n = 27 b - 1                                   . . . . . 2
Now,
m - n = 0 ⇒ 9 b + 8 - 27 b - 1 = 0                             Using   1   and   2 ⇒ 9 b - 27 b + 8 + 1 = 0 ⇒ - 18 b + 9 = 0 ⇒ - 18 b = - 9 ⇒ b = - 9 - 18 = 1 2
Thus, the value of b is 1 2 .

Page No 56:

Question 13:

Answer:


8 m 2 + 3 m - 6 - 9 m - 7 + 3 m 2 - 2 m + 4 = 8 m 2 + 3 m - 6 - 9 m + 7 + 3 m 2 - 2 m + 4 = 8 m 2 + 3 m 2 + 3 m - 9 m - 2 m - 6 + 7 + 4 = 11 m 2 - 8 m + 5

Page No 56:

Question 14:

Which polynomial is to be subtracted from x 2 + 13x + 7 to get the polynomial 3x 2 + 5x - 4?

Answer:


Let p(x) be the polynomial which is to be subtracted from x 2 + 13x + 7 to get the polynomial 3x 2 + 5x &minus 4.

Thus, the required polynomial is &minus2x 2 + 8x + 11.

Page No 56:

Question 15:

Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?

Answer:


The required polynomial can be obtained by subtracting the polynomial 4m + 2n + 3 from 6m + 3n + 10.

Thus, the polynomial 2m + n + 7 is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10.


6.6: Divide Polynomials

Dividing

Honestly, this is definitely a tough subject. It can be hard to understand, so we will start slow and work up to it.

Dividing a polynomial by a monomial (one term) is a good place to start because it’s not that bad. Let’s say we have:

To be able to divide in this example, we need to divide each piece by 3x and reduce.

It becomes much tougher if we have to divide by a binomial (two terms). If this is the case, we have to do long division. Now, you’ll have to reach back in your memory to try to remember how to do long division. Let’s work though a basic example.

Seven does not fit into 2, so we have to see how many times it will fit into 24 without going over. Seven times three works!

egin
-21 & \
hline
& 35
end
(bring down the 5)

Is this ringing a bell at all, yet? Let’s finish.

Let’s try to apply this to a polynomial.

You only have to worry about the 2x to start with. What can we multiply 2x by to match 2x 3 ? The answer is x 2 .

Now we want our 2x to match -2x 2 . We can multiply by –x.

6x and 25 cannot be combined since they are not “like terms,” so we leave it as 6x + 25. Almost done! To get 2x to match -6x we multiply by -3.

There is no way to get our 2x to match 7 by multiplying so we leave the last piece of the answer in fraction form or (Large frac<7><<2x - 6>>).

So our final answer is: ( - x - 3 + Large frac<7><<2x - 6>>).

Phew! That is a tough problem! Let’s try one more.

Final answer: ( - 5k - 1 + Large frac<6><<2k - 5>>)

Below you can download some free math worksheets and practice.


Source code

You can download the source of the current program and the old sum polynomial factorization applet from GitHub. Notice that the source code is in C language and you need the Emscripten environment in order to generate Javascript.

Written by Dario Alpern. Last updated 1 July 2021.

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Watch the video: Section Dividing Polynomials (October 2021).