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RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 6 - Factorisation of Polynomials.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.1

Solution 1


Solution 2


Solution 3


Solution 4

Solution 5

Solution 6

Solution 7

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  .  

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.2

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

Solution 2


Solution 3


Solution 4

Solution 5

Solution 6

Solution 7

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7


Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6


Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13


Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

         Let p(x) = x3 + 13x2 + 32x + 20
         The factors of 20 are 1, 2, 4, 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

                      We know that
         Dividend = Divisor  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)

Solution 12

        Let p(x) = x3 - 3x2 - 9x - 5
        Factors of 5 are 1,  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

                   Now, Dividend = Divisor  Quotient + Remainder
         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)  

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,                      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)

   

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.34

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).

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).

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).

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).

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).

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).

Solution 7

begin mathsize 12px style When space straight a space polynomial space straight p left parenthesis straight x right parenthesis space is space divided space by space straight q left parenthesis straight x right parenthesis space straight i. straight e. space left parenthesis straight x space plus-or-minus straight alpha right parenthesis space then space straight p left parenthesis minus-or-plus space straight alpha right parenthesis space is space the space remainder.
If space straight x space plus-or-minus space straight alpha space is space the space factor space of space polynomial comma space then space remainder space is space apostrophe 0 apostrophe.
So space if space straight x to the power of 51 space plus space 51 space is space divided space by space straight x space plus space 1 comma
remainder space equals space left parenthesis negative 1 right parenthesis to the power of 51 space plus space 51 space equals negative 1 space plus space 51 space equals space 50
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

 

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).

Solution 9

begin mathsize 12px style straight x space plus space straight a space is space straight a space factor space of space polynomial comma space straight p left parenthesis straight x right parenthesis space equals space straight x to the power of 4 space end exponent minus space straight a squared straight x squared space plus space 3 straight x space minus space 6 straight a space
Then comma space at space straight x equals negative straight a comma space straight p left parenthesis straight x right parenthesis space equals space 0 space
rightwards double arrow left parenthesis negative straight a right parenthesis to the power of 4 space minus space straight a squared left parenthesis negative straight a right parenthesis squared space plus space 3 left parenthesis negative straight a right parenthesis space minus space 6 straight a space equals space 0
rightwards double arrow up diagonal strike straight a to the power of 4 end strike space minus space up diagonal strike straight a to the power of 4 end strike space minus space 3 straight a space minus space 6 straight a space equals space 0
rightwards double arrow negative 9 straight a space equals space 0
rightwards double arrow straight a space equals space 0
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

 

Solution 10

begin mathsize 12px style Let space straight p left parenthesis straight x right parenthesis space equals space 4 straight x cubed space plus space 3 straight x squared space minus space 4 straight x space plus space straight k
Now comma space if space left parenthesis straight x space minus space 1 right parenthesis space is space straight a space facor space of space straight p left parenthesis straight x right parenthesis comma space then space at space straight x space equals space 1 comma space straight p left parenthesis straight x right parenthesis equals space 0
So comma space straight p left parenthesis 1 right parenthesis space equals space 0
rightwards double arrow space 4 left parenthesis 1 right parenthesis cubed space plus space 3 left parenthesis 1 right parenthesis squared space minus space 4 left parenthesis 1 right parenthesis space plus space straight k space equals space 0
rightwards double arrow space up diagonal strike 4 space plus space 3 space minus up diagonal strike space 4 end strike space plus space straight k space equals space 0
rightwards double arrow space straight k space equals space minus 3
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

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).

RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.35

Solution 12

begin mathsize 11px style If space straight f left parenthesis straight x right parenthesis space is space straight a space polynomial space and space straight f left parenthesis straight alpha right parenthesis space equals space 0 comma space then space left parenthesis straight x space minus straight alpha right parenthesis space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space or
vice space versa space if space left parenthesis straight x space minus space straight alpha right parenthesis space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space then space straight f left parenthesis straight alpha right parenthesis space equals space 0.
Now comma space straight f open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses equals 0
So comma space at space straight x space equals space fraction numerator negative 1 over denominator 2 end fraction comma space straight f open parentheses straight x close parentheses space equals space 0
or space at space 2 straight x space equals space minus 1 comma space straight f open parentheses straight x close parentheses space equals space 0
or space at space 2 straight x space plus space 1 space equals space 0 comma space straight f left parenthesis straight x right parenthesis space equals space 0
rightwards double arrow open parentheses 2 straight x space plus space 1 close parentheses space is space straight a space factor space of space straight f open parentheses straight x close parentheses
Hence comma space straight c orrect space option space is space left parenthesis straight b right parenthesis. end style

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).

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).

Solution 15

If space straight x space minus space 1 space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space then space definitely space straight f left parenthesis 1 right parenthesis space equals space 0
And comma space straight x space minus space 1 space is space not space straight a space factor space of space straight g left parenthesis straight x right parenthesis comma space then space straight g left parenthesis 1 right parenthesis space not equal to space 0 space space
So comma space at space straight x space equals space 1 space space
option space left parenthesis straight a right parenthesis space straight f left parenthesis 1 right parenthesis space straight g left parenthesis 1 right parenthesis space equals space 0 space cross times space straight g left parenthesis 1 right parenthesis equals space 0
option space left parenthesis straight b right parenthesis space minus straight f left parenthesis 1 right parenthesis space plus space straight g left parenthesis 1 right parenthesis space equals space 0 space plus space straight g left parenthesis 1 right parenthesis space equals space straight g left parenthesis 1 right parenthesis space not equal to space 0
option space left parenthesis straight c right parenthesis space straight f left parenthesis 1 right parenthesis space minus space straight g left parenthesis 1 right parenthesis space equals space 0 space minus space straight g left parenthesis 1 right parenthesis space equals space minus straight g left parenthesis 1 right parenthesis space not equal to space 0
option space left parenthesis straight d right parenthesis space left curly bracket straight f left parenthesis 1 right parenthesis space plus space straight g left parenthesis 1 right parenthesis right curly bracket space straight g left parenthesis 1 right parenthesis space equals space left curly bracket 0 space plus space straight g left parenthesis 1 right parenthesis right curly bracket space straight g left parenthesis 1 right parenthesis equals left curly bracket straight g left parenthesis 1 right parenthesis right curly bracket squared space not equal to space 0
So space at space straight x space equals space 1 space only comma space straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space equals space 0
Thus comma space left parenthesis straight x space minus space 1 right parenthesis space is space factor space of space straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space too.
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. space space

 

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).

Solution 17

begin mathsize 12px style Let space straight p open parentheses straight x close parentheses space equals space 3 straight x cubed space plus space 8 straight x squared space plus space 8 straight x space plus space 3 space plus space 5 straight k space and space straight q open parentheses straight x close parentheses space equals space straight x squared space plus space straight x space plus space 1
Now comma space if space straight q open parentheses straight x close parentheses space is space straight a space factor space of space straight p open parentheses straight x close parentheses comma space then space arranging space straight p open parentheses straight x close parentheses space in space order space to space have space straight q open parentheses straight x close parentheses space in space common comma
straight p open parentheses straight x close parentheses space equals space 3 straight x cubed space plus space 3 straight x squared space plus space 3 straight x space plus space 5 straight x squared space plus space 5 straight x space plus space 3 space plus space 2 space minus space 2 space plus space 5 straight k space space space space space space left square bracket adding space plus 2 comma space minus 2 space in space straight p left parenthesis straight x right parenthesis right square bracket
space space space space space space space space space equals 3 straight x left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis space plus space 5 left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis space plus space 5 straight k space minus space 2
straight p left parenthesis straight x right parenthesis space equals space left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis left parenthesis 3 straight x space plus space 5 right parenthesis space plus space 5 straight k space minus space 2 space space space space space.... left parenthesis 1 right parenthesis
From space equation space left parenthesis 1 right parenthesis comma space we space can space see space if space we space divide space straight p left parenthesis straight x right parenthesis space by space straight q left parenthesis straight x right parenthesis comma
then space quotient space will space be space left parenthesis 3 straight x space plus space 5 right parenthesis space and space remainder space will space be space left parenthesis 5 straight k space minus space 2 right parenthesis
But space straight q space left parenthesis straight x right parenthesis space is space straight a space factor space of space straight p left parenthesis straight x right parenthesis.
So comma space remainder space equals space 0 space rightwards double arrow space 5 straight k space minus space 2 space equals space 0 space space rightwards double arrow space straight k equals 2 over 5
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

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).

Solution 19

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis space equals space px squared space plus space 5 straight x space plus space straight r
Now comma space if space straight x space minus space 2 space and space straight x space minus 1 half space are space factors space of space straight f left parenthesis straight x right parenthesis comma
then space at space straight x space equals space 2 space and space straight x equals 1 half comma space straight f left parenthesis straight x right parenthesis space equals space 0.
So comma space straight f left parenthesis 2 right parenthesis space equals space 0 comma space space straight f open parentheses 1 half close parentheses equals 0
rightwards double arrow straight p open parentheses 2 close parentheses squared space plus space 5 open parentheses 2 close parentheses space plus space straight r space equals space 0 space space space space space space space space space space space space space space space and space space space space space space space space straight p open parentheses 1 half close parentheses squared plus 5 open parentheses 1 half close parentheses plus straight r space equals space 0
rightwards double arrow 4 straight p space plus space straight r space plus space 10 space equals space 0 space space space space.... left parenthesis 1 right parenthesis space space space space space space space space and space space space space space space space space space 4 straight r space plus space straight p space plus space 10 space equals space 0 space space space space space space.... left parenthesis 2 right parenthesis
Subtracting space equation space left parenthesis 2 right parenthesis space from space left parenthesis 1 right parenthesis comma space we space have
3 straight p space minus space 3 straight r space equals space 0
rightwards double arrow straight p space equals space straight r
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 20

begin mathsize 11px style If space straight x squared space minus space 1 space is space factor space of space straight p left parenthesis straight x right parenthesis space equals space ax to the power of 4 space plus space bx cubed space plus space cx squared space plus space dx space plus space straight e comma
then space left parenthesis straight x space minus space 1 right parenthesis space and space left parenthesis straight x space plus space 1 right parenthesis space will space also space be space factors space of space straight p left parenthesis straight x right parenthesis.
straight B ecause space straight x squared space minus space 1 space equals space left parenthesis straight x space minus space 1 right parenthesis left parenthesis straight x space plus space 1 right parenthesis
straight T hen comma space at space straight x space equals space 1 space and space straight x space equals space minus 1 comma space straight p left parenthesis straight x right parenthesis space equals space 0
rightwards double arrow space straight p left parenthesis 1 right parenthesis space equals space 0 space and space straight p left parenthesis negative 1 right parenthesis space equals 0
rightwards double arrow space straight a space plus space straight b space plus space straight c space plus space straight d space plus space straight e space equals space 0 space space space.... left parenthesis 1 right parenthesis space space space space space space space space space space and space space space space space space straight a space minus space straight b space plus space straight c space minus space straight d space plus space straight e space equals space 0 space space space space.... left parenthesis 2 right parenthesis
Adding space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma
2 straight a space plus space 2 straight c space plus space 2 straight e space equals space 0
rightwards double arrow space straight a space plus space straight c space plus space straight e space equals space 0 space space space space.... left parenthesis 3 right parenthesis
Substracting space equation space left parenthesis 2 right parenthesis space from space left parenthesis 1 right parenthesis
2 straight b space plus space 2 straight d space equals space 0
rightwards double arrow straight b space plus space straight d space equals space 0 space space space.... left parenthesis 4 right parenthesis
From space equations space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space get space
straight a space plus space straight c space plus space straight e space equals space straight b space plus space straight d
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 6 - Factorisation of Polynomials for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 6 - Factorisation of Polynomials.

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CBSE IX - Mathematics

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