RD SHARMA Solutions for Class 9 Maths Chapter 6 - Factorisation of Polynomials

Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.1

Question 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 8
Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution 8
Degree of a polynomial is the highest power of variable in the polynomial.
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .  
Monomial has only one term in it. So monomial of degree 100 can be written as 7x100.
 
Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable.  The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with  .  

Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.2

Question 1
If f(x) = 2x3 - 13x2 + 17x + 12, find:

(i) f(2)

(ii) f(-3)

(iii) f(0)
Solution 1
(i)

f(x) = 2x3 - 13x2 + 17x + 12

f(2) = 2(2)3 - 13(2)2 + 17(2) + 12

      = 16 - 52 + 34 + 12

      = 10

(ii)

f(-3) = 2(-3)3 - 13(-3)2 + 17(-3) + 12

       = -54 - 117 - 51 + 12

       = - 210

(iii)

f(0) = 2(0)3 - 13(0)2 + 17(0) + 12

      = 0 - 0 + 0 + 12
   
      =12
Question 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 7
Find rational roots of the polynomial f(x) = 2x3 + x2 - 7x - 6.
Solution 7
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.3

Question 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 7
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 7
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 8
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 8
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 9
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 9
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 10

If the polynomials ax3 + 3x2 - 13 and 2x3 -5x + a, when divided by (x-2) leave the same remainder, find the value of a.

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 11
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
 
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 11

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.4

Question 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 4
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 5
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 6
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 7
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 7
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 8
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 8
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 9
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 9
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 10
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 10
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 11
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 11
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 12
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 13
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 13
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 14
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 14
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 15
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 15
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 16

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 16

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 20

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 20

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 21

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 21

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 22

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 22

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 23

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 23

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 24

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 24

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 25

What must be added to 3x3 + x2 - 22x + 9 so that the result is exactly divisible by 3x2 + 7x - 6?

Solution 25

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.5

Question 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 1
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 2
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Solution 3
Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
Question 4

Using factor theorem, factorize: x4 - 7x3 + 9x2 + 7x -10

Solution 4

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 5

Using factor theorem, factorize: 3x3 - x2 - 3x + 1

Solution 5

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 6

Using factor theorem, factorize each of the following polynomials:

x3 - 23x2 + 142x - 120

Solution 6

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 7

Using factor theorem, factorize: y3 - 7y + 6

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 8

Using factor theorem, factorize: x3 - 10x2 - 53x - 42

Solution 8

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 9

Using factor theorem, factorize: y3 - 2y2 - 29y - 42

Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 10

Using factor theorem, factorize: 2y3 - 5y2 - 19y + 42

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 11

x3 + 13x2 + 32x + 20

Solution 11

         Let p(x) = x3 + 13x2 + 32x + 20
         The factors of 20 are Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials1, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 2, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 4, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 5 ... ...
         By hit and trial method
         p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
                   = - 1 + 13 - 32 + 20
                   = 33 - 33 = 0
         As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).

         Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
          By long division

 
         Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
 
         We know that
         Dividend = Divisor Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials Quotient + Remainder
         x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
                                         = (x + 1) (x2 + 10x + 2x + 20)
                                         = (x + 1) [x (x + 10) + 2 (x + 10)]
                                         = (x + 1) (x + 10) (x + 2)
                                         = (x + 1) (x + 2) (x + 10)
Question 12

Factorise:

x3 - 3x2 - 9x - 5

Solution 12

        Let p(x) = x3 - 3x2 - 9x - 5
        Factors of 5 are Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials1, Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials 5.
        By hit and trial method
        p(- 1) = (- 1)3 - 3(- 1)2 - 9(- 1) - 5
           = - 1 - 3 + 9 - 5 = 0
        So x + 1 is a factor of this polynomial
        Let us find the quotient while dividing x3 + 3x2 - 9x - 5 by x + 1
        By long division

 
       Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
 
        Now, Dividend = Divisor Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials Quotient + Remainder
        Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0
                                     = (x + 1) (x2 - 5 x + x - 5)
                                     = (x + 1) [(x (x - 5) +1 (x - 5)]
                                     = (x + 1) (x - 5) (x + 1)
                                     = (x - 5) (x + 1) (x + 1)
 
Question 13

Factorise:

2y3 + y2 - 2y - 1

Solution 13

         Let p(y) = 2y3 + y2 - 2y - 1

         By hit and trial method
         p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
                = 2 + 1 - 2 - 1= 0
         So, y - 1 is a factor of this polynomial
         By long division method,
         Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials
 
          p(y) = 2y3 + y2 - 2y - 1
                 = (y - 1) (2y2 +3y + 1)
                 = (y - 1) (2y2 +2y + y +1)
                 = (y - 1) [2y (y + 1) + 1 (y + 1)]
                 = (y - 1) (y + 1) (2y + 1)

 
 
Question 14

Using factor theorem, factorize: x3 - 2x2 - x + 2

Solution 14

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 15

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 15

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 16

Using factor theorem, factroize : x4 - 2x3 - 7x2 + 8x + 12

Solution 16

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 17

Using factor theorem, factroize : x4 + 10x3 + 35x2 + 50x + 24

Solution 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 18

Using factor theorem, factorize : 2x4 - 7x3 - 13x2 + 63x - 45

Solution 18

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Chapter 6 - Factorisation of Polynomials Exercise 6.34

Question 1

If x - 2 is a factor of x2 + 3ax - 2a, then a =

(a) 2

(b) -2

(c) 1

(d) -1

Solution 1

Let p(x) = x2 + 3ax - 2a be the given polynomial.

x - 2 is a factor of p(x).

Thus, p(2) = 0

(2)2 + 3a × 2 - 2a = 0

4 + 4a = 0

a = -1

Hence, correct option is (d).

Question 2

If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =

(a) -6

(b) -7

(c) -8

(d) -10

Solution 2

Since, p(x) = x3 + 6x2 + 4x + k is exactly divisible by x + 2,

(x + 2) is a factor of p(x).

So, p(-2) = 0

i.e (-2)3 + 6(-2)2 + 4(-2) + k = 0

     -8 + 24 - 8 + k = 0

     24 - 16 + k = 0

     8 + k = 0

     k = -8

Hence, correct option is (c).

Question 3

If x - a is a factor of x3 - 3x2a + 2a2x + b, then the value of b is

(a) 0

(b) 2

(c) 1

(d) 3

Solution 3

Let p(x) = x3 - 3x2a + 2a2x + b

(x - a) is a factor of p(x). 

So, p(a) = 0

a3 - 3a2.a + 2a2.a + b = 0

a3 - 3a3 + 2a3 + b = 0

3a3 - 3a3 + b = 0

b = 0

Hence, correct option is (a).

Question 4

If x140 + 2x151 + k is divisible by x + 1, then the value of k is

(a) 1

(b) -3

(c) 2

(d) -2

Solution 4

Let p(x) = x140 + 2x151 + k

Since p(x) is divisible by (x + 1),  

(x + 1) is a factor of p(x).

So, p(-1) = 0

(-1)140 + 2(-1)151 + k = 0

1 + 2(-1) + k = 0

1 - 2 + k = 0

k - 1 = 0

k = 1

Hence, correct option is (a).

Question 5

If x + 2 is a factor of x2 + mx + 14, then m =

(a) 7

(b) 2

(c) 9

(d) 14

Solution 5

If x + 2 is a factor of x2 + mx + 14,

then at x = -2,

x2 + mx + 14 = 0

i.e. (-2)2 + m(-2) + 14 = 0

4 - 2m + 14 = 0

2m = 18

m = 9

Hence, correct option is (c).

Question 6

If x - 3 is a factor of x2 - ax - 15, then a =

(a) -2

(b) 5

(c) -5

(d) 3

Solution 6

x - 3 is a factor of x2 - ax - 15,

then at x = 3,

x2 - ax - 15 = 0

i.e. (3)2 - a(3) - 15 = 0

9 - 3a - 15 = 0

a = -2

Hence, correct option is (a).

Question 7

If x51 + 51 is divided by x + 1, the remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Solution 7

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

 

Question 8

If x + 1 is a factor of the polynomial 2x2 + kx, then k =

(a) -2

(b) -3

(c) 4

(d) 2

Solution 8

x + 1 is a factor of p(x) = 2x2 + kx

Then, p(-1) = 0

i.e. 2(-1)2 + k(-1) = 0

2 - k = 0

k = 2

Hence, correct option is (d).

Question 9

If x + α is a factor of x4 - a2x2 + 3x - 6a, then a =

(a) 0

(b) -1

(c) 1

(d) 2

Solution 9

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

 

Question 10

The value of k for which x - 1 is a factor of 4x3 + 3x2 - 4x + k, is

(a) 3

(b) 1

(c) -2

(d) -3

Solution 10

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 11

If x + 2 and x - 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively

(a) 5 and -3

(b) 17 and -8

(c) 7 and -18

(d) 23 and -19

Solution 11

If (x + 2) and (x - 1) are factors of polynomial x3 + 10x2 + mx + n,

then x = -2, x = +1 will satisfy the polynomial.

Let p(x) = x3 + 10x2 + mx + n

Then, p(-2) = 0

(-2)3 + 10(-2)2 + m(-2) + n = 0

-8 + 40 - 2m + n = 0

32 - 2m + n = 0            ....(1)

And, p(1) = 0

(1)3 + 10(1)2 + m(1) + n = 0

1 + 10 + m + n = 0

11 + m + n = 0                   ....(2)

Substracting equation (1) from equation (2), we get

-21 + 3m = 0

3m = 21

m = 7       

Substituting m = 7 in equation (2),

11 + 7 + n = 0

18 + n = 0

n = -18

Hence, correct option is (c).

Chapter 6 - Factorisation of Polynomials Exercise 6.35

Question 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 12

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 13

When x3 - 2x2 + ax - b is divided by x2 - 2x - 3, then remainder is x - 6. The values of a and b are respectively

(a) -2, -6

(b) 2 and -6

(c) -2 and 6

(d) 2 and 6

Solution 13

Let p(x) = x3 - 2x2 + ax - b, r(x) = x - 6 and q(x) = x2 - 2x - 3

Then q(x) is a factor of [p(x) - r(x)] 

{because if p(x) is divided by q(x), remainder is r(x). So, [p(x) - r(x)] will be exactly divided by q(x)}

Now, q(x) = x2 - 2x - 3 = (x - 3)(x + 1)

If q(x) is a factor of [p(x) - r(x)] then (x - 3) and (x + 1) are also factors of [p(x) - r(x)]

So, at x = 3 and x = -1, p(x) - r(x) will be zero.

Now p(3) - r(3) = 0

i.e. (3)3 - 2(3)2 + a(3) - b - (3 - 6) = 0

i.e. 27 - 18 + 3a - b + 3 = 0

i.e. 3a - b + 12 = 0  ....(1)

And, p(-1) - r(-1) = 0

i.e. (-1)3 - 2(-1)2 + a(-1) - b - (-1 - 6) = 0

i.e. -1 - 2 - a - b + 7 = 0

i.e -a - b + 4 = 0    ....(2)

Subtracting equation (2) from equation (1), we get

4a + 8 = 0

a = -2

From (2), -(-2) - b + 4 = 0

b = 6

Hence, correct option is (c).

Question 14

One factor of x4 + x2 - 20 is x2 + 5. The other factor is

(a) x2 - 4

(b) x - 4

(c) x2 - 5

(d) x + 2

Solution 14

x4 + x2 - 20

= x4 + 5x2 - 4x2 - 20

=x2(x2 + 5) - 4(x2 + 5)

= (x2 + 5)(x2 - 4)

So, other factor is x2 - 4.

Hence, correct option is (a).

Question 15

If (x - 1) is a factor of polynomial f(x) but not of g(x), then it must be a factor of

(a) f(x) g(x)

(b) -f(x) + g(x)

(c) f(x) - g(x)

(d) {f(x) + g(x)} g(x)

Solution 15

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

 

Question 16

(x + 1) is a factor of xn + 1 only if

(a) n is an odd integer

(b) n is an even integer

(c) n is a negative integer

(d) n is a positive integer

Solution 16

If x + 1 is a factor of xn + 1,

then, at x = -1, xn + 1 = 0

(-1)n + 1 = 0

(-1)n = -1

(-1)n will be equal to -1 if and only if n is an odd integer.

If n is even, then (-1)n = 1

So, n should be an odd integer.

Hence, correct option is (a).

Question 17

If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + 3 + 5k, then the value of k is

(a) 0

(b) 2/5

(c) 5/2

(d) -1

Solution 17

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 18

If (3x - 1)7 = a7x7 + a6x6 + a5x5 + ... + a1x + a0, then a7 + a6 + a5 + ... + a1 + a0 =

(a) 0

(b) 1

(c) 128

(d) 64

Solution 18

Correct option (c)

(3x - 1)7 = a7x7 + a6x6 + ......... + a1x + a0  ....(1)

Putting x = 1 in equation (1), we have

[3(1) - 1]7 = a7 + a6 + ..... + a1 + a0

So, a7 + a6 + a5 + ..... + a1 + a0 = 2= 128

Hence, correct option is (c).

Question 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Solution 19

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials

Question 20

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e, then

(a)  a + c + e = b + d

(b) a + b + e = c + d

(c) a + b + c = d + e

(d) b + c + d = a + e

Solution 20

Rd-sharma Solutions Cbse Class 9 Mathematics Chapter - Factorisation Of Polynomials