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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

CHAPTER

2We are Starting from a Point but want to Make it a Circle of Infinite Radius

Polynomial

Constant: - A symbol having a fixed numerical value is called a constant.

Example: = ,7

5,6,8 . etc.

Variables: A symbol which may be assigned different numerical values is known as variable.

Example: In equation 5x = 0, 5 is constant while x is variable.

Algebraic Expressions: -

A combination of constant and variable, connected by some or all of the operations, + ,

-, and , is known as an algebraic expression.

Terms of an algebraic Expressions:-

The several parts of an algebraic expression separated by + or – operations are called the terms of the

expression.

Example: x3 + 2x

2y - 4xy

2 + y

3 + t, is an algebraic expression containing 5 terms, namely, x3, 2x

2y,

xy2, y

3 and 7.

Polynomials:-

An algebraic expression is which the variables involved have only non-negative integral powers is

called a polynomial.

Example: 3x3 – 2x

2 + 4x – 3 a polynomial is one variable x.

Condition of an Algebraic Expression to be a

(i) Exponent should be a whole number; it should not be a negative integer.

(ii) Exponent should not be in fraction.

(iii) Variable should not be under radical sign.

Coefficients:

In polynomial 3x3 – 2x

2 + 5x – 4, we say that the coefficient of x

3, x

2 and x are 3, – 2 and 5

respectively and also, – 4 is the constant term.

Degree of a polynomial: -

The highest power of the variable is called the degree of polynomial.

Example:

(i) 2x + 3, is a polynomial is x of degree 1.

(ii) yy2

34 2 is a polynomial is y of degree 2.

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

Polynomial of Various Degrees-:

(i) Linear Polynomial: - A polynomial of degree 1 is called a linear polynomial.

Example: 3x + 5

2y – 9

(ii) Quadratic Polynomial:- A polynomial of degree 2 is called a quadratic polynomial.

Example: 2

152

xx

583 2 yy

Cubic Polynomial: - A polynomial of degree 3 is called is a cubic polynomial

Example:

(i) 4x3 – 3x

2 + 7x + 1

(ii) 8x3 – 2x

2 – 5x – 8

Biquadratic Polynomial: - A polynomial of degree 4 is called a biquadratic polynomial

Example:

(i) x4 – 3x

3 + 2x

2 + 5x – 3

(ii) 4x4 – 7x

3 + 6x

2 + 9x – 7

Number of Terms:

(i) Monomial: - A polynomial containing one non zero term is called a monomial

Example: - 5, 3x, xy2

1are called monomial.

(ii) Binomial:- A polynomial containing two non-zero terms is called binomial.

Example: - 3 + 6x, 2x2 + 5x, are binomial.

(iii) Trinomial: - A polynomial containing three non-zero terms is called a trinomial.

Example: - 3x2 – 2x + 1, 2y

2 – 1 + 2y are trinomial.

Constant polynomial: - A polynomial containing one term only, consisting of a constant is called a

constant is called a constant polynomial.

Examples: 3, - 5, 5

4 are all constant polynomial.

Note: (i) Every real number is a constant polynomial.

(ii) The degree of a non-zero constant polynomial is zero.

Zero Polynomial: - A polynomial consisting of one term, namely zero only, is called a zero

polynomial.

(i) The degree of a zero polynomial is not defined.

(ii) A polynomial is one variable x of degree n is an expression of the form

011

1 .......... axaxaxa nn

nn

where a0, a1, a2 ……….. are constant.

Standard form of a polynomial:-

A polynomial is written in the descending powers in x is called standard form of a polynomial

Example: x3 – 3x

2 + 2x + 1 is a polynomial is the standard form as the powers in descending order

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IX ACADEMIC QUESTIONS Subjective

Assignment – 1

1. Which of the following expression are

polynomials in one variable and which are

not? State reason for your answer

In case of a polynomial, write its degree.

(i) 4x2 – 3x + 7

(ii) 2

1

1

112

xx

(iii) )2()1(1

xxx

(iv) )64(8

1 2 xx

(v) 322 y

(vi) xx 23

(vii) 50312 tyx

(viii) 1

(ix) 222

1 2 xx

(x) )1(

)1()1( 2

x

xxx

2. Write the degree of each of the following

polynomials:

(i) 83 1059 xxx

(ii) 3

(iii) 5 t

(iv) 12- x + 2x3

(v) 100x + 1

(vi) 52 x

(vii) xxx 745 23

(viii) xx 2

2

(ix) 12 x

(x) 1143 3 yy

3. Write the coefficient of: -

(i) yyisy 332

(ii) x is 2x2 + x

(iii) 3x is 122 2 xx

(iv) x5 is 3x – 6x

2 – 2x

2 + 3x

5

(v) x2 is )64(

8

1 2 xx

(vi) x is 2

11112

xx

(vii) r2 is r

2

(viii) t is 3 + t 2

(ix) xy4 is 6y

3 + 9y

4 + 10x

2 – 8y +

5

2

(x) x3 is x

2 (x – 1)

4. Give examples: -

(i) Binomial of degree 35

(ii) Monomial of degree 100

(iii) Trinomial of degree 27

(iv) Binomial of degree 74

(v) Monomial of degree 16.

5. Classify the following as linear, quadratic

and cubic polynomials: -

(i) 2x2 + 4x

(ii) P3

(iii) 2x – y – y2

(iv) x – x3

(v) 5t

(vi) – 7 + z

(vii) r2

(viii) y + y2 + 4

(ix) 1 + x

(x) x3 + 3x

6. Rewrite the following in standard form

(i) x

xxx

1

)1()1( 2

(ii) x – 7 + 8x2 + 9x

3

(iii) – 5x2 + 6 – 3x

3 + 4x

(iv) – 4 + 6x3 – x + 7x

4

22x

(v) y2 + 5y

3 – 11– 7y + 9y

4

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

Zero of a Polynomial

Let p (x) be a polynomial. If P (x) = 0, then we say that is a zero of the polynomial p (x).

The value of the variable (s) is the equation which may satisfy the equation, is caused, the zero of the

polynomial.

Value of a polynomial:- The value of a polynomial f (x) at x = is obtained by substituting x = is the

given polynomial and is denoted by f (x).

Note: (i) The zero of the polynomial is also called the root of the polynomial.

(ii) A non-zero constant polynomial has no zero.

(iii) Every real number is a zero of the zero polynomial.

Important Points

A zero of polynomial need to be zero.

0 may be a zero of a polynomial.

Every linear polynomial has one and only one zero.

A polynomial can have more than one zero.

Number of zeroes is same as the degree of polynomial.

Example: Find the value of the polynomial at 5x + 4x2 + 3 at

(i) x = 0 (ii) x = - 1 (iii) x = 2

Answer: (i) p(x) = 5x + 4x2 + 3 => p(0) = 5(0) + 4(0)

2 + 3 = 3

(ii) p(x) = 5x + 4x2 + 3 => p(-1) = 5(-1) + 4(-1)

2 + 3 => 5 - 4(1) + 3 = -6

(iii) p(x) = 5x + 4x2 + 3 => p(2) = 5(2) + 4(2)

2 + 3 => 10 - 16 + 3 = -3

Example: Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = -1/3

(ii) p(x) = 5x - π, x = 4/5

(iii) p(x) = x2 - 1, x = 1, -1

Answer: (i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.

At, p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0

Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

(ii) If x = 4/5 is a zero of polynomial p(x) = 5x - π then p(4/5) should be 0.

At, p(4/5) = 5(4/5) - π = 4 - π

Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x - π.

(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x2 - 1, then p(1) and p(-1) should be 0.

At, p(1) = (1)2 - 1 = 0 and

At, p(-1) = (-1)2 - 1 = 0

Hence, x = 1 and -1 are zeroes of the polynomial p(x) = x2 - 1.

Example: Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5

(ii) p(x) = x - 5

(iii) p(x) = 2x + 5

Answer :(i) p(x) = x + 5 =>p(x) = 0 =>x + 5 = 0 => x = -5

Therefore, x = -5 is a zero of polynomial p(x) = x + 5

(ii) p(x) = x - 5 => p(x) = 0 => x - 5 = 0 => x = 5

Therefore, x = 5 is a zero of polynomial p(x) = x - 5.

(iii) p(x) = 2x + 5 => p(x) = 02x + 5 = 0 => 2x = -5 =>x = -5/2

Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5

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011-26925013/14+91-9811134008+91-9582231489

NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IX ACADEMIC QUESTIONS Subjective

Assignment - 2

1. Find the value of the polynomial f (x) 2x3 – 13x

2 + 17 x + 12 at:

(i) f (2) (ii) f (-3) (iii) f = 0

2. Find P (0), p (1) and p (2) for the following polynomials:-

(i) P (x) = 4x2 + x – 5 (ii) P (z) = (z + 1) ( z – 1)

(iii) P (y) = 9y3 + 2y

2 + y + 7 (iv) P (t) = t

4 + t + 1

3. If x = 0 and x = -1 are the roots of the polynomial f (x) = 2x3 – 3x

2 + ax + b, find the value of a and b

4. If x = 2 is a zero of the polynomial f (x) = 2x2 – 3x + 7 a, find the value of a.

5. Find the zero (s) of the following polynomial: -

(i) P (x) = ax + b, a 0

(ii) h (x) = 6x – 1

(iii) f (x) = 5x – 7

(iv) b(x) = ( x + 1) ( x – 1 )

(v) p (x) = x + 3

(vi) p (t) = 2t – 3

(vii) 2x + 1

(viii) 2x + 0

(ix) x – 5

(x) x13

24

6. Verify whether the indicate numbers are zeroes of the polynomial corresponding to them in

following cases: -

(i) p (x) = 2x + 1, x = 2

1

(ii) p (x) = x2, x = 0

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

(iii) p (x) = 5 – 4x, x = 5

4

(iv) q (x) = 4x, x = - 4

(v) q (x) = bx – b, x = 1

7. Verify that 2 and – 3 are the zeroes of the polynomial p (x) = x2 + x – 6.

Remainder Theorem:

Let f (x) be a polynomial of degree 1n and let be any real number.

When f (x) is divided by (x – a), then the remainder is f (a).

If p (x) and g (x) are two polynomial such that degree of p (x) degree of g (x) and g (x)

0, then we can find polynomials q (x) and r (x), such that: p (x) = g (x) q (x) + r (x) where r (x) = 0

or degree of r (x) < degree of g (x). p (x) is divided by 9 (x), gives q (x) as quotient and r (x) as

remainder.

Factor Theorem

Let p (x) be any polynomial of degree greater than or equal to 1 and ‘a’ be any real number, then

(i) (x – a) is a factor of p (x) if P (a) = 0

(ii) p (a) = 0 if (x – a) is a factor of p (x).

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IX ACADEMIC QUESTIONS Subjective

Assignment - 3 (Remainder Theorem)

1. Divide x + 3x2 – 1

2. Find the remainder obtained on dividing p(x) = x3 + 1 by g (x) = x + 1

3. Find the remainder when p (x) = 4x3 – 12x

2 + 14x – 3 is dividing by g (x) =

2

1x

4. Divide the p (x) = x3 – ax

2 + 6x – a by x – a

5. If the polynomial ax3 + 4x

2 + 3x – 4 and x

3 – 4x + a leave the same remainder when divided by

(x – 3) find the value of a.

6. Let R1 and R2 are the remainders when the polynomials x3 + 2x

2 – 5ax - 7 and x

3

+ ax2 – 12x + 6 are divided by x + 1 and x – 2 respectively. If 2R1 + R2 = 6, find the value of a.

7. Check whether the polynomial f (x) = 4x3 + 4x

2 – x – 1 is a multiple of 2x + 1.

8. The polynomial kx2 + 3x

3 + 6 when divided by x – 2. leaves a remainder which is double the

remainder left by the polynomial 2x3 + 17x + k when divided by x – 2, find the value of k.

9. Divide- 3x4 + 2x

3 –

27

2

3

2

x

by 3

2x

10. If f (x) = x4 – 2x

3 + 3x

2 – ax + b is a polynomial such that when it is divided by (x- 1) and (x + 1)

the remainders are respectively 5 and 19. Determine the values of a and b.

(Factor Theorem)

1. Use factor theorem to determine whether 2x is a factor of 2522 2 xx

2. Find the value of k, if (x- 3) is factor of k2x

2 – kx – 2.

3. Show that:-

(i) (x + 5) is a factor of (2x3 + 9x

2 – 11x – 30)

(ii) )2( is a factor of (7x2 - 24 x – 6)

(iii) (x – 3) is a factor of (2x3 + 7x

2 – 24x – 45)

(iv) (x2 + 2x – 3) is a factor of x

3 – 3x

2 – 13x + 15

4. Show that 3x3 + x

2 – 20x + 12 is exactly divisible by 3x – 2.

5. For what value of m is the p (x) = 2x4 + 3x

3 + 2mx

2 + 3x + 6 exactly divisible by (x +2)?

Factors: A polynomial g (x) is called a factor of the polynomial p (x) if g (x) divides the p (x)

exactly.

Example: - (x – 2) is a factor of (x2 + 3x – 10)

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

Factorization: To express a given polynomial as the product of polynomials, each of degreeless

than that of the given polynomial such that no such factor has a factor of lower degree, is called

factorization.

Example: (x2 – 16) can be expressed as (x – 4 ) (x + 4) also (x

2 – 3x + 2) = (x – 2) (x – 1)

Methods of Factorization

1. Factorization by taking out the common factor:-

When each term of an expression has a common factor, we divide each term by this factor and take it

out as a multiple.

Example:

(i) 5x2 – 20 xy

5 is common in each term of the given expression.

So, 5 (x2 – 4xy), now, also, x is common

so, the factorization is : 5x (x – 4)

so, 5x2 – 20xy = 5x (x – 4y)

2. Factorization by grouping: -

Sometimes in a given expression it is not possible to take out a common factor directly. However,

the term of the given expression are grouped in such a manner that we may have a common factor.

This can now be factorize by the method of taking out a common factor.

Example: -

(i) 6 ab – b2 + 12ac – 2 bc

We can group the expression as follows: 6ab + 12ac – b2 – 2bc

now, by taking out a common factor as: 6a (b + 2c) – b (b + 2 c)

so, by taking (b + 2c) common, (b + 2c) (6a – b)

so, p (x) can be factories as : p (x) = (b + 2c ) (6a – b)

3. Factorization of Quadratic Trinomials

Case 1: - Polynomial of the form x2 + bx + c. We find integers p and q such that

p + q = b and pq = c. then, x2 + bx + c = x

2 + (p + q) x + pq

= x2 + px + qx + pq

= x (x + p) + q (x + p)

= (x + p) (x + q)

Example: - p (x) = x2 + 9x + 18

We try to split q into two parts whose sum is 9 and product 18.

so, 6 + 3 = 9 and 6 3 = 18

x2 + 9x +18 = x

2 + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

Hence, x2 + 9x + 18 = (x + 6) (x + 3)

Case 2 : Polynomial of the form ax2 + bx + c

In this case, we find integers p and q such that

p + q = b and pq = ac. Then,

ax2 + bx + c =ax

2 + (p + q)x

a

pq

= a2x

2 + apx + aqx + pq

= ax (ax + p) q (ax + p)

= (ax + p) (x + q)

Hence, (ax2 + bx + c) = (ax + p) (ax + q)

Example:

p (x) = 6x2 + 7x – 3

Here, 6 ( – 3) = – 18

so, we try to split 7 into two parts whose sum is 7 and product – 18.

Clearly, 9 + (-2) = 7 and 9 (– 2) = – 18

6x2 + 7x – 3 = 6x

2 + 9x – 2x – 3

= 3x (2x + 3) – (2x + 3)

= (3x – 1) (2x + 3)

Hence, (6x2 + 7x – 3) = (2x +3) (3x – 1)

x2 + 9x + 18 = x

2 + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

Hence, x2 + 9x + 18 = (x + 6) (x + 3)

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011-26925013/14+91-9811134008+91-9582231489

NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IX ACADEMIC QUESTIONS Subjective

Assignment - 4

1. Factorize the given expressions: -

(1) a2 + b – ab – a

(2) ab + bc + ax + cx

(3) 6x2 + 5x – 6

(4) y2 – 5y + 6

(5) 6x2 + 17x + 5

(6) x2 + 3x + x + 3

(7) 6(2a + 3b)2 – 8 (2a + 3b)

(8) 5a (b + c) – 7b (b + c)

(9) x2 – 4x – 21

(10) 6x2 + 7x – 3

(11) x (x – y)3 + 3x

2y (x – y)

(12) x3 – 23x

2 + 14 2x – 120

(13) x3 + 6x

2 + 11x + 6

(14) 2y3 – 5y

2 – 19y + 42

(15) x3 + 13x

2 + 32x + 20

(16) 4x3 + 4x

2 – x – 1

(17) x3 – 6x

2 + 3x + 10

(18) 6x3 + 11x

2 – 3x – 2

(19) x3 – x

2 – 9x + 9

(20) 2y3 + y

2 – 2y – 1

2. Without actual division prove that 2x4 – 6x

3 + 3x

2 + 3x – 2 is exactly divisible by x

2–3x+2.

3. Find the value of a, if x – a is a factor of x3 – x

2x + x + 2.

4. Factorise the following by the method of middle term splitting:

(i) 36113 2 xx (ii) 32534 2 xx (iii) 9x2 – 22x + 8

(iv) 2x2 – 7x – 39 (v)

12

1

6

52 2

xx

5. Show that (x – 1) is a factor of x 10

– 1 and also of x11

– 1

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IDENTITIES FOR FACTORIZATION

(i) (x + y)2 = x

2 + 2xy + y

2

Ex. 9x2 + 6xy + y

2

Ans: 9x2 + 6xy + y

2 = (3x)

2 + (2×3x×y) + y

2 = (3x + y)

2 =

(3x + y) (3x + y)= (3x + y)

2

(ii) (x – y)2 = x

2 – 2xy + y

2

Ex. 4y2 - 4y + 1

Ans: 4y2 - 4y + 1 = (2y)

2 - (2×2y×1) + 1

2 = (2y - 1)

2 =

(2y - 1) (2y - 1)= (2y - 1)

2

(iii) x2 – y

2 = (x + y) ( x – y)

Ex. 104 × 96

Ans: 104 × 96 = (100 + 4) (100 - 4) = (100)2 - (4)

2 = 10000 - 16 = 9984

(iv) (x + a) (x + b) = x2 + (a + b) x + ab

Ex. (x + 4) (x + 10)

Ans: In (x + 4) (x + 10), a = 4 and b = 10

Now, (x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10) => x

2 + 14x

+ 40

(v) (x + y + z)2 = x

2 + y

2 + z

2 + 2xy + 2yz + 2zx

(vi) (x + y)3 = x

3 + y

3 + 3xy (x + y)

(vii) (x – y)3 = x

3 – y

3 – 3xy (x – y)

(viii) x3 + y

3 + z

3 – 3xyz = (x + y + z)

(x2 + y

2 + z

2 –xy - yz – zx)

37

011-26925013/14+91-9811134008+91-9582231489

NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

IX ACADEMIC QUESTIONS Subjective

Assignment - 5

Question 1 - Factorize:

1. 25x2 – 64y

2

2. 2a5 – 32a

3. x4 – 625

4. 108 a2 – 3 (b – c)

2

5. (a + b)3 – a – b

6. 27a2 – 48 b

2

7. 2 – 50x2

8. (3x + 5y)2 – 4z

2

9. 4x2 + 9y

2 + 16z

2 + 12xy – 24yz – 16xz

10. 9x2 + 16y

2 + 4z

2 – 24xy + 16yz – 12xz

11. 2 2 22x y 8z 2 2 xy 4 2 yz 8xz

12. 25p2 + 4q

2 + 9r

2 – 20pq – 12qr + 30 pr

13. x6 – y

6

14. a3 - 22 b

3

15. 2a3 – 128 a

16. 3

3

27

18

yx

17. abccba 21827822 333

18. a3 – b

3 + 1 + 3ab

19. a3 – 8b

3 + 64c

3 + 24abc.

20. (2x + 3y)3 – (2x – 3y)3

Question 2 – Find the Product

(i) ( x + y – z) (x2 + y

2 + z

2 – zy + yz + zx)

(ii) (x–2y + 3) (x2+4y

2 + 2xy – 3x + 6y + 9)

(iii) (a– b – c) (a2 + b

2 + c

2 + ab + ac – bc)

3. If (x + y + z) = 0, prove that

(x3 + y

3 + z

3) = 3xyz

4. x + y + 4 = 0, find the value of (x3 + y

3 – 12xy + 64)

5. Evaluate by using identities:-

(i) (106)3

(ii) 105 102

(iii) (997)2

(iv) (103) (97)

(v) 95 96

(vi) (x + 8) (x – 10 )

(vii) (999)3

(viii) (x – 3) (x + 5)

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CLASS - IXMathematics

Polynomial

(ix) (105)3

(x) (2x – 4 )(2x + 4 )

6. If P = 2 – a, prove that

a3 + 6ap + p

3 – 8 = 0.

7. Expand:

(i) (3x + 2)3

(ii) 3

25

4

x

(iii) 3

13

2

x

(iv) (3a – 2ab)3

(v) (2a + 3b + 4c)

(vi) (3a – 5b – c)2

(vii) 2

24

1

2

1

ba

8. Verify: -

(i) x3 + y

3 = (x + y) (x

2 – xy + y

2 )

(ii) x3 – y

3 = (x – y) (x

2 + xy + y

2 )

(iii) x3 – y

3 = (x – y) (x

2 + xy + y

2 )

9. Verify that : x

3 + y

3 – 3xyz =

2

1

10. (x + y + z) 222 )()()( xzzyyx

11. Without actually calculating the cubes, find the value of each of the following:

(i) (– 12)3 + (7)

3 + (5)

3 (ii) (28)

3 + (– 15)

3 + (– 13)

3

12. If x = 2y + 6, find the value of (x3 – 8y

3 – 36 xy – 216) ]

13. Without actual division, prove that (2x4 – 6x

3 + 3x

2 + 3x – 2) is exactly divisible by (x

2 –

3x + 2)

14. If polynomials (2x3 + ax

2 + 3x – 5) and (x

3 + x

2 – 2x + a) leave the same remainder when divided by

(x – 2), find the value of a. Also, find the remainder in each case.

15. The polynomial ax3 + 3x

2 – 13 and 2x

3 – 5x + a are divided by x + 2. If the remainder in each case in

the same, find the value of a.

16. If f (x) = x4 – 2x

3 + 3x

2 – ax + b is a polynomial such that when it is divided by (x –

1) (x + 1), the remainders are respectively 5 and 19. Determine the remainder when f (x) is

divided by ( x – 2)

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

17. Find the remainder when f (x) is divided by g (x) and verify the result by actual division.

a. f (x)=9x3 – 3x

2 + x – 5, g (x) =

3

2x

b. f (x) = 3x4 + 2x

3

3

2x

27

2

9

x, g (x) =

3

2x

c. f (x) = x3 + 4x

2 – 3x + 10, g (x) = x + 4

18. The polynomials ax3 + 3x

2 – 3 and 2x

3 – 5x + a when divided by (x – 4) leave the remainders R1 and

R2 respectively. Find the values of a is each of the following cases, if:

(i) R1 = R2 (ii) R1 + R2 = 0 (iii) 2R1 – R2 = 0

19. Find the values of a and b so that the polynomial x3 + 10x

2 + ax + b is exactly divisible x – 1 as well as

x – 2.

20. If both x – 2 and 2

1x are factors of px

2 + 5x + r show that p = r

21. If x2 – 1 is a factor of ax

4 + bx

3 + cx

2 + dx + e, show that a + c + e = b + d = 0

22. Use factor theorem to verify that x + a is a factor of x2 + –a

2 for any positive integer.

23. Verily whether 0 and 3 are the zeroes of the polynomial p (x) = x2 – 3x.

24. Verify that – 1/2 is a zeroes of p (y) = 2 y + 1

25. If p (x) = (x – a) q (x) then prove that.

The degree of q (x) = the degree of p (x) – 1

26. Using factor theorem, show that x – y, y – z, z – x, are the factor of x (xy2 – z

2) + y (z

2 – x

2) + z (x

2 – y

2)

27. Factories the p (x) = x4 + x

2y

2 + y

4

28. Factories: 2

2

2

2

2a

b

b

a

29. Factories: a2 – b

2 – 4c

2 + 4d

2 – 4 (ad – bc)

30. Evaluate: 2 2[(999) (1) ]

31. Factories: 4x2 – z

2 + 9y

2 – 4 p

2 + 4pz – 12xy

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

XI SCIENCE & DIP. ENTRANCE Multiple Choice Questions

Assignment – 6

1. A symbol which may be assigned different numerical value is known as

(a) Constant (b) variable (c) Expression (d) None of these

2. An Algebraic expression in which the variables involved have only non-negative integral powers is

called

(a) Algebraic identities (b) Term (c) Polynomial (d) All of these

3. The parts of an algebraic expression are called

(a) Operation (b) Factors (c) Both a & b (d) Term

4. A combination of constants and variable connected by some or all of the operation is known as

(a) Coefficient (b) Degree (c) Zero (d) None of these

5. In a polynomial, exponent should be a

(a) Negative integer (b) Fraction (c) Whole number (d) Rational

6. Variable in a polynomial should not be under : -

(a) Radical sign (b) Negative sign, (c) Both a and b (d) None of these

7. The coefficient of x2 is equation 2x

2 + 5x – 4 is:

(a) 0 (b) 2 (c) 5 (d) 4

8. The coefficient of x in 1 is -

(a) 1 (b) 2 (c) 6

(d) None of these

9. The highest power of the variable is called the ______ of polynomial.

(a) Term (b) Coefficient (c) Operation (d) Degree

10. The degree of x in 2 is: -

(a) 10 (b) 2 (c) 0 (d) All of these

11. Linear polynomial is a polynomial of degree: -

(a) 100 (b) 1 (c) 0 (d) 5

12. The degree of constant polynomial is always: -

(a) 0 (b) 1 (c) 2 (d) 4

13. A polynomial of degree 2 is called a: -

(a) Cubic polynomial (b) Constant

(c) Biquadratic polynomial (d) Quadratic polynomial

14. Biquadratic polynomial is a polynomial of degree

(a) 2 (b) 4 (c) 16 (d) Both b & c

15. Monomial is a polynomial containing _____ non-zero term:

(a) Two (b) Zero (c) One (d) Four

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

16. Cubic polynomial is a polynomial of degree.

(a) 3 (b) 5 (c) 0 (d) 25

17. A polynomial containing one term only, consisting the variable to degree O is called.

(a) Constant polynomial (b) Monomial (c) Binomial (d) Trinomial

18. A polynomial containing three non – zero terms is: -

(a) Binomial (b) Constant polynomial (c) Trinomial (d) Coefficient.

19. Binomial is a polynomial containing _______ non-zero term is called: -

(a) Four (b) Three (c) Six (d) Two

20. –5 is an example of ________ polynomial:

(a) Monomial (b) Constant (c) Linear (d) both a & b

21. The degree of a non – zero constant polynomial is : -

(a) 1 (b) 2 (c) 0 (d) 4

22. A polynomial consisting of one term namely zero only is called a _________ polynomial : -

(a) Linear (b) Quadratic (c) Cubic (d) Zero

23. The degree of a zero polynomial is : -

(a) 0 (b) 1 (c) Not defined (d) None of these

24. Which of the following is the standard form of polynomial: -

(a) 1 – x + x2 + x

3 (b) x

2 + x

3 + x

4 – 0 (c) x

3 + x

2–x + 1 (d) None of these

25. Tick out the odd one: -

(a) 1 (b) 2

11112

xx (c) 4x

2 – 3x+7 (d) xx 23

26. Which of the polynomial has degree 0:

(a) 0 (b) 5 (c) x2 – 1 (d) x

27. Among the following, which is a monomials of degree 100:

(a) x100

– x90

+ 1 (b) 2x100

(c) 0.x100

(d) All of these

28. The value of variable in the equation which may satisfy the equation, is called the ___________ of

polynomial: -

(a) Coefficient (b) Zero (c) Both a & b (d) None of these

29. The zero of a polynomial is called _________ of polynomial.

(a) Real (b) Root (c) Coefficient (d) All of these

30. A non-zero constant polynomial has _____ zero (s): -

(a) Two (b) One (c) No (d) Both a & b

31. Every real number is a zero of the ______ p (x):

(a) Constant (b) Monomial (c) Zero (d) All of these

32. Every _________ polynomial has only one zero: -

(a) Cubic (b) Quadratic (c) Both a & b (d) linear

33. Number of zeroes is same as the _______ of p (x):

(a) Zero (b) Coefficient (c) Both a & b (d) None of these

34. (x – a) is a factor of p (x) is: -

(a) 0)( ap (b) 0)( ap

(c) 0)( ap (d) 0)( ap

35. Product of (x + y + z ) (x2 + y

2 + z

2 – xy –yz – zx) =

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011-26925013/14+91-9811134008+91-9582231489

NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

(a) (x2 + y

2 + z

2 – 3xyz (b) x

3 – y

3 – z

3 + 3xyz

(c) x3 – y

3 + z

3 – 3xyz (d) x

3 + y

3 + z

3 – 3xyz

36. x2 + 2xy + y

2 =

(a) x2 + y

2 (b) x

2 + y

2 + 2 (c) (x + y)

2 (d) (x – y)

2

37. If x + y + z = 0, then x3 + y

3 + z

3 =

(a) 3zyz

(b) (x + y + z)3

(c) x2 + y

2 + z

2 (d) 2xyz ‘

38. (105)3 can be expanded as: -

(a) (100 + 5)3

(b) (110 – 5)3 (c) Both are correct (d) None

39. 95 96 can be expanded as

(a) (90 + 5) (100 – 4) (b) (90 + 5) (90 + 6)

(c) (90 + 5) 96 (d) All of these

40. (2)3 + (– 1)

3 + (–1)

3 is equal to

(a) 8

(b) (2)3

(c) 6 (d) (– 1)3

41. x2 + y

2 + z

2 + 2xy + 2yz + 2zx is =

(a) (2xy + 2zx + 2yz)2

(b) (x2 – y

2 –z

2)

(c) (z + y + x) (d) None of these

42. (x + a) (x + b) is same as:

(a) x2 – (– a – b) x + ab (b) x

2 + (a + b) x + ab

(c) Both a & b (d) None of these

43. The zero of p (x) = x – 3 is

(a) – 3

(b) 0 (c) 3 (d) None of these

44. The zero of p (x) = 2x – 1 is

(a) 2

(b) – 2 (c) – 1/2 (d) 1/2

45. The zero of p (x) = x is

(a) 1

(b) 0 (c) 10 (d) All are correct

46. x6 – y

6 can be solved by using identity: -

(a) x2 – y

2 (b) x

3 – y

3 (c) Both a & b (d) None of these

47. 0.x100

is a:-

(a) Monomial

(b) Binomial (c) Zero polynomial (d) constant

48. Coefficient of t is 23 t is : -

(a) 3

(b) 2

(c) t (d) 23

49. Coefficient of x3 is x

2 (x – 1) is

(a) 0

(b) 2 (c) 1 (d) – 1

50. Degree of t in p (t) = 5t is

(a) 0

(b) 5 (c) 1 (d) 2

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

ANSWERS

Assignment – 1

1. (i) yes, since the variable power is non negative integer and degree = 2.

(ii) No (iii) No

(iv) yes, since the variable power is non negative integer and degree = 2.

(v) yes, since the variable power is non negative integer and degree = 2.

(vi) No

(vii) yes, since the variable power is non negative integer and degree = 50

(viii) yes, since the variable power is non negative integer and degree = 0.

(ix) yes, since the variable power is non negative integer and degree = 2.

(x) No

2. (i) 10 (ii) 0 (iii) 1 (iv) 3 (v) 1 (vi) 1 (vii) 3

(viii) 2 (ix) 1 (x) 3

3. (i) 0 (ii) 1 (iii) 0 (iv) 3 (v) 1/8 (vi) 1 (vii) 1

(viii) 2 (ix) 0 (x) 1

4. (i) x35

+ x (ii) y100

(iii) x27

+ x26

+ x (iv) 2x74

– 1 (v) 7z16

5. (i) Quadratic (ii) Cubic (iii) Quadratic (iv) Cubic (v) Cubic (vi) Linear

(vii) Linear (viii) Quadratic (ix) Linear (x) Cubic

6. (i) x2 + x + 1 (ii) 9x

3 + 8x

2 + x – 7 (iii) –3x

3 – 5x

2 + 4x + 6

(iv) 7x4 + 6x

3 – 2 x

2 – x – 4 (v) 9y

4 + 5y

3 + y

2 – 7y – 11

Assignment – 2

1. (i) 10 (ii) – 108 (iii) 12

2. (i) p(0) = –5, p(1) = 0, p(2) = 13 (ii) p(0) = –1, p(1) = 0, p(2) = 3

(iii) p(0) = 7, p(1) = 19, p(1) = 89 (iv) p(0) = 1, p(1) = 3, p(3) = 19

3. a = 5, b = 0

4. a = –2/7

5. (i) –b/a (ii) 1/6 (iii) 7/5 (iv) 1, –1 (v) –3 (vi) 3/2

(vii) –1/2 (viii) x = 0 (ix) x = 5 (x) 13/24

6. (i) No (ii) Yes (iii) No (iv) No (v) Yes

7. 2 is not but –3 is the root of p(x).

Assignment – 3 : (Remainder Theorem)

1) 1 2) 0 3) 3 4) a = 0 5) a = –1 6) a = 2

7) yes, f(x) is the multiple of 2x + 1. 8) k = 33 9) 0 (10) a = 5, b = 8

(Factor Theorem)

1. Yes x 2 is the factor 2. 2 1

K or K3 3

3. m = –1

Assignment – 4 : 1. Factorize

1. (a – 1)(a – b) 2. (x + b)(a + c) 3. (2x + 3)(3x – 2) 4. (y – 2)(y – 3)

5. (2x + 5) (3x + 1) 6. (x + 1) (x + 3) 7. 2(2a + 3b) (6a + 9b – 4) 8. (b + c) (5a – 7b)

9. (x + 3 (x – 7) 10. (2x + 3) (3x – 1) 11. x(x – y) (x2 + xy + y

2)

12. (x – 1) (x – 10) (x – 12) 13. (x + 1)(x + 2)(x + 3) 14. (y – 2)(y + 3) (2y – 7)

15. (x + 1)(x + 2)(x + 10) 16. (2x – 1)(2x + 1)(x + 1) 17. (x + 1)(x – 2) (x – 5)

18. (3x + 1) (2x – 1) (x + 2) 19. (x + 3)(x – 3)(x – 1) 20. (y – 1) (y + 1) (2y + 1)

3. a = –2

4. (i) (x 3 3)( 3x 2) (ii) ( 3x 2)(4x 3) (iii) (x 2)(9x 4) (iv) (x + 3)(2x – 13)

(iv) (x + 3) (2x – 13 (v) (6x – 1)(4x – 1)

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NTSE, NSO Diploma, XI Entrance

CLASS - IXMathematics

Polynomial

Assignment – 5

1. Factorize :

1. (5x + 8y) (5x – 8y) 2. 2a(a2 + 4) (a + 2) (a – 2) 3. (x

2 + 25) (x + 5)(x – 5)

4. 3((6a + b – c) (6a – b + c) 5. (a + b)(a + b + 1)(a + b – 1) 6. 3(3a + 4b)(3a – 4b)

7. 2(a – 5x)(a + 5x) 8. (3x + 5y + 2z)(3x + 5y – 2z)

9. (2x + 3y – 4z)(2x + 3y – 4z) 10. (–3x + 4y + 2z)(–3x + 4y + 2z)

11. ( 2x y 2 2z) 12. (5p – 2q + 3r) (5p – 2q + 3r)

13. (x – y)(x + y)(x2 + 2b

2 + 2 ab 14.

2 22b)(a 2b 2ab)(a

15. 2a(a + 8)(a – 8) 16. 22

1 1 2x2x 4x

3y 3y9y

17. 2 2 2( 2a 2b 3c)(2a 4b ac 2 2ab 6bc 3 2ac)

18. 2 2(a b 1)(a b 1 ab b a)

19. (a – 2b + 4c) ( a 2 + 4b

2 + 16c

2 + 2ab + 8bc – 4ac) 20.6y(4x

2 + 9y

2)

2. Find the Product:

(i) x3 + y

3 – z

3 + 3xyz (ii) x

3 – 8y

2 + 27 + 18xy (iii) a

3 – b

3 – c

3 – 3abc

4. 0

5. Evaluate

(i) 119106 (ii) 10710 (iii) 994009 (iv) 9991 (v) 9120

(vi) x2 – 2x – 80 (vii) 997002999 (viii) x

2 + 2x – 15 (ix) 1157625 (x) 4x

2 – 16

7. Expand:

(i) 27x3 + 8 + 54x

2 + 36x (ii) 3 264 96 48

x 8 x x125 25 5

(iii) 3 28 4x 1 x 2x

27 3

(iv)

23

3 2

1 27a 9a27a

4b64b 16b (v) 4a

2 + ab

2 + 16c

2 + 12ab + 24bc + 16ac

(vi) 9a2 + 25b

2 + c

2 – 30ab + 10bc – 6ac (vii) 2 21 1 1

a b 4 ab b 2a4 16 4

10. (i) –1180 (ii) 16380 11. 0 13. a = –3 and remainder = 5

14. a = 5/9 15. 7

16. (i) –3 (ii) 0 (iii) 92

17. (i) a = 1 (ii) a = –153/65 (iii) a = 18/127

18. a = –37, b = 26 26. (x2 + y

2 + xy)(x

2 + y

2 – xy) 27.

2a b

b a

28. (a + b – 2c – 2d) (a – b + 2c – 2b) 29. 998000

30 (2x – 3y + 2p – z) (2x – 3y – 2p + z)

Assignment – 6 1. b 2.c 3.b 4.d 5.c 6.c 7.b 8.d 9.d 10.c 11.b 12.a

13.d 14.b 15.c 16.a 17.b 18.c 19.d 20.d 21.c 22.d 23.c 24.d

25.d 26.b 27.b 28.b 29.b 30.c 31.a 32.d 33.d 34.d 35.d 36.c

37.a 38.c 39.b 40.c 41.d 42.c 43.c 44.d 45.b 46.c 47.c 48.b

49.c 50.a