# ML Aggarwal Solutions for Chapter 6 Factorization Class 10 Maths ICSE

Here, we are providing the solutions for Chapter 6 Factorization from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the sixth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 6 Factorization ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on the simplifying algebraic expressions, factorizing quadratic polynomial and factorizing cubic polynomials.

1. Find the remainder (without divisions) on dividing f(x) by x – 2, where

(i) f(x) = 5x2 – 1x + 4

(ii) f(x) = 2x3 – 7x2 + 3

Let x – 2 = 0, then x = 2

(i) Substituting value of x in f(x)

f(x) = 5x2 – 7x + 4

⇒ f(2) = 5(2)2 – 7(2) + 4

⇒ f(2) = 20 – 14 + 4 = 10

Hence Remainder = 10

(ii) f(x) = 2x3 – 7x2 + 3

∴ f(2) = 2(2)3  – 7(2)2  + 3

= 16 – 28 + 3

Hence remainder = - 9

2. using remainder theorem, find the remainder on dividing f(x) by (x + 3) where

(i) f(x) = 2x2 – 5x + 1

(ii) f(x) = 3x3 + 7x2 – 5x + 1

Let x + 3 = 0

⇒ x = - 3

Substituting the value of x in f(x),

(i) f(x) = 2x2 – 5x + 1

∴ f(-3) = 2(-3)2 – 5(-3) + 1

= 18 + 15 + 1

= 34

Hence Remainder = 34

(ii) f(x) = 3x3 + 7x2 – 5x + 1

= 3(-3)3 + 7(-3)2 – 5(-3) + 1

= - 81 + 63 + 15 + 1

= - 2

Hence Remainder = - 2

3. Find the remainder (without division) on dividing f(x) by (2x + 1) where

(i) f(x) = 4x2 + 5x + 3

(ii) f(x) = 3x2 – 7x2 + 4x + 11

Let 2x + 1 = 0, then x = - (1/2)

Substituting the value of x in f(x):

(i) f(x) = 4x2 + 5x + 3

= 4(- 1/2)2 + 5 × (-1/2) + 3

= 4 × 1/4 – 5/2 +3

= 1 – 5/2 + 3

= 4 – 5/2

= 3/2

∴ Remainder = 3/2

(ii) f(x) = 3x3 – 7x2 + 4x + 11

= - 3(- 1/2)3 – 7(- 1/2)2 + 4(- 1/2) + 11

= 3(- 1/8) – 7(1/4) + 4(- 1/2) + 11

= - 3/8 – 7/4 – 2 + 11

= (- 3 – 14 – 16 + 88)/8

= 55/8

= 6.7/8

4. (i) Find the remainder (without division) when 2x3 – 3x2 + 7x – 8 is divided by x – 1

(ii) Find the remainder (without division) on dividing 3x2 + 5x – 9 by (3x + 2)

(i) Let x – 1 = 0, then x = 1

Substituting value of x in f(x)

f(x) = 2x3 – 3x2 + 7x – 8

= 2(1)3 – 3(1)2 + 7(1) – 8

= 2 × 1 – 3 × 1 + 7 × 1 – 8

= 2 – 3 + 7 – 8

= - 2

∴ Remainder = 2

(ii) Let 3x + 2 = 0, then 3x = - 2

⇒ x = -2/3

Substituting the value of x in f(x)

f(x) = 3x2 + 5x – 9

= 3(-2/3)2 + 5{-(2/3)} – 9

= 3 × 4/9 – 5 × 2/3 – 9

= 4/3 – 10/3 – 9

= - 6/3 – 9

= - 2 – 9

= - 11

∴ Remainder = - 11

5. Using remainder theorem, find the value of k if on dividing 2x2 + 3x2 – kx + 5 by x – 2, leaves a remainder 7.

f(x) = 2x2 + 3x2 – kx + 5

g(x) = x – 2, if x – 2 = 0, then x = 2

Dividing f(x) by g(x) the remainder will be

f(2) = 2(2)3 + 3(2)2 – k × 2 + 5

= 16 + 12 – 2k + 5

= 33 – 2k

Remainder = 7

∴ 33 – 2k = 0

⇒ 33 – 7 = 2k

⇒ 2k = 26

⇒ k = 26/2 = 13

∴ k = 13

6. Using remainder theorem, find the value of a if the division of x3 + 5x2 – ax + 6 by (x – 1) leaves the remainder 2a.

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x)

f(x) = x3 + 5x2 – ax + 6

= (1)3 + 5(1)2 – a(1) + 6

= 1 + 5 – a + 6

= 12 – a

∵ Remainder = 2a

∴ 12 – a = 2a

⇒ 12 = a + 2a

⇒ 3a = 12

∴ a = 4

7. (i) What number must be subtracted from 2x2 – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?

(ii) What number must be added to 2x3 – 7x2 + 2x so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3?

(i) Let a be subtracted from 2x2 – 5x,

Dividing 2x2 – 5x by 2x + 1,

Here remainder is (3 – a) but we are given that remainder is 2.

∴ 3 – a = 2

⇒ - a = 2 – 3 = - 1

⇒ a = 1

Hence, 1 is to be subtracted.

(ii) Let a be added to 2x3 – 7x2 + 2x dividing by 2x – 3, then

But remainder is – 2, then

a – 6 = - 2

⇒ a = - 2 + 6

⇒ a = 4

Hence, 4 is to be added.

8. (i) When divided by x – 3 the polynomials x2 – px2 + x + 6 and 2x3 – x2 – (p + 3)x – 6 leave the same remainder. Find the value of ‘p’

(ii) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3.

(i) By dividing x3 – px2 + x + 6

And 2x3 – x2 – (p + 3)x – 6

By x – 3, the remainder is same Let x – 3 = 0, then x = 3

Now by Remainder Theorem,

Let p(x) = x3 – px2 + x + 6

p(3) = (3)3 – p(3)2 + 3 + 6

= 27 – 9p + 9

= 36 – 9p

and q(x) = 2x3 – x2 – (p + 3)x – 6

q(3) = 2(3)2 – (3)2 – (p + 3) × 3 – 6

= 2 × 27 – 9 – 3p – 9 – 6

= 54 – 24 – 3p – 9 – 6

= 54 – 24 – 3p

= 30 – 3p

∵ The remainder in each case is same

∴ 36 – 9p = 30 – 3p

36 – 30 = 9p – 3p

⇒ 6 = 6p

⇒ p = 6/6 = 1

∴ p = 1

(ii) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3.

The given polynomials are ax3 + 3x2 – 9 and 2x3 + 4x + a

Let p(x) = ax3 + 3x2 – 9

and q(x)= 2x3 + 4x + a

Given that p(x) and q(x) leave the same remainder when divided by (x + 3),

Thus by Remainder Theorem, we have

p(-3) = q(-3)

⇒ a(-3)3 + 3(-3)2 – 9 = 2(-3)3 + 4(-3) + a

⇒- 27a + 27 – 9 = - 54 – 12 + a

⇒ - 27a – a = - 66 – 18

⇒ - 28a = - 84

⇒ a = 84/28

∴ a = 3

9. By factor theorem show that (x + 3) and (2x – 1) are factors of 2x2 + 5x – 3.

Let x + 3 = 0 then x = - 3

Substituting the value of x in f(x)

f(x) = 2x2 + 5x – 3 = 2(- 3)2 + 5(- 3) – 3

f(- 3) = 18 – 15 – 3 = 0

∵ Remainder = 0, then x + 3 is a factor

Again let 2x – 1 = 0, then x = 1/2

Substituting the value of x in f(x),

f(x) = 2x2 + 5x – 3

f(1/2) = 2(1/2)2 + 5(1/2) – 3

= 2 × 1/4 + 5/2 – 3

= 1/2 + 5/2 – 3 = 0

∵ Remainder = 0,

∴ 2x – 1 is also a factor

Hence, proved

10. Without actual division, prove that x4 + 2x3 – 2x2 + 2x – 3 is exactly divisible by x2 + 2x – 3.

x2 + 2x – 3

= x2 + 3x – x – 3

= x(x + 3) – 1(x + 3)

= (x + 3)(x – 1)

Let p(x) = x4 + 2x3 – 2x2 + 2x – 3

We see that

p(-3) = (- 3)4 + 2(- 3)3 – 2(-3)2 + 2(-3) – 3

= 81 – 54 – 18 – 6 – 3

= 0

Hence by converse of factor theorem, (x + 3) is a factor of p(x).

Also we see that

p(1) = (1)4 + 2(1)3 - 2(1)2 + 2(1) – 3 = 0

Hence by converse of factor theorem, (x – 1) is a factor of p(x).

From above, we see that

(x + 3)(x – 1), i.e., x2 + 2x – 3 is a factor of p(x)

⇒ p(x) is exactly divisible by (x2 + 2x – 3)

11. Show that (x – 2) is a factor of 3x2 – x – 10. Hence factorise 3x2 – x – 10.

Let x – 2 = 0, then x = 2

Substituting the value of x in f(x),

f(x) = 3x2 – x – 10

= 3(2)2 – 2 – 10

= 12 – 2 – 10 = 0

∵ Remainder is zero

∴ x – 2 is a factor of f(x).

Dividing 3x2 – x – 10 by x – 2, we get

∴ 3x2 – x – 10 = (x – 2)(3x + 5)

12. (i) Show that (x – 1) is a factor of x3 – 5x2 – x + 5. Hence, factorise x3 – 5x2 – x + 5.

(ii) Show that (x – 3) is a factor of x3 – 7x2 + 15x – 9. Hence, factorise x3 – 7x2 + 15x – 9

(i) Let x – 1 = 0, then x = 1

Substituting the value of x in f(x),

f(x) = x3 – 5x2 – x + 5

= (1)3 – 5(1)2 – 1 + 5

= 1 – 5 – 1 + 5

= 0

∵ Remainder = 0

∴ x – 1 is a factor of x3 – 5x2 – x + 5

Now dividing f(x) by x – 1, we get ∴ x3 – 4x2 – x + 5

= (x – 1)(x2 – 4x – 5) = (x – 1)[x2 – 5x + x – 5]

= (x – 1)[x(x – 5) + 1(x – 5)]

= (x – 1)(x + 1)(x – 5)

(ii) Let x – 3 = 0, then x = 3,

Substituting the value of x in f(x), f(x) = x3 – 7x2 + 15x – 9

= (3)3 – 7(3)2 + 15(3) – 9

= 27 – 63 + 45 – 9

= 72 – 72

= 0

∵ Remainder = 0,

∴ x – 3 is a factor of x3 – 7x2 + 15x – 9

Now dividing it by x – 3, we get

∴ x3 – 7x2 + 15x – 9

= (x – 3)(x2 – 4x + 3) = (x – 3)[x2 – x – 3x + 3]

= (x – 3)[x(x – 1) – 3(x – 1)]

= (x – 3)(x – 1)(x – 3) = (x – 3)2(x – 1)

13. Show that (2x + 1) is a factor of 4x3  + 12x2  + 11x + 3. Hence factorise 4x3 + 12x3  + 11x + 3.

Let 2x + 1 = 0,

Then x = - 1/2

Substituting the value of x in f(x),

f(x) = 4x3 + 12x2 + 11x + 3

f(-1/2) = 4(-1/2)3 + 12(-1/2)2 + 11(-1/2) + 3

= 4(- 1/8)

= - 1/2 + 3 – 11/2 + 3

= (6) – (6) = 0

∵ Remainder = 0,

∴ 2x + 1 is a factor of

4x3 + 12x2 + 11x + 3

Now dividing f(x) by 2x + 1, we get

∴  4x3 + 12x2 + 11x + 3

= (2x + 1)(2x2 + 5x + 3)

= (2x + 1)[2x2 + 2x + 3x + 3]

(2x + 1)[2x(x + 1) + 3(x + 1)]

= (2x + 1)[(x + 1)(2x + 3)]

= (2x + 1)(x + 1)(2x + 3)

14. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence factorise the given expression completely, using the factor theorem.

Let 2x + 7 = 0, then 2x = - 7

x = (-7)/2

Substituting the value of x in f(x),

f(x) = 2x3 + 5x2 – 11x – 14

f(- 7/2) = 2(- 7/2)3 + 5(-7/2)2 – 11(- 7/2) – 14

= (-343)/4 + 245/4 + 77/2 – 14

= (- 343 = 245 + 154 – 56)/4

= (- 399 + 399)/4

= 0

Hence, (2x + 7) is a factor of f(x).

Now, 2x3 + 5x2 – 11x – 14 = (2x + 7)(x2 – x – 2)

= (2x + 7)[x2 – 2x + x – 2]

= (2x + 7)[x(x – 2) + 1(x – 2)]

= (2x + 7)(x + 1)(x – 2)

15. Use factor theorem to factorise the following polynomials completely.

(i) x3 + 2x2 – 5x – 6

(ii) x3 – 13x – 12

(i) Let f(x) = x3 + 2x2 – 5x – 6

Factors of (∵ 6 = ± 1; ± 2, ± 3, ± 6)

Let x = - 1, then

f(-1) = (-1)3 + 2(-1)2 – 5(-1) – 6

= - 1 + 2(1) + 5 – 6

= - 1 + 2 + 5 – 6

= 7 – 7

= 0

∵ f(-1) = 0

∴ x + 1 is a factor of f(x)

Now, dividing f(x) by x + 1, we get

f(x) = (x + 1) (x2 + x – 6)

= (x + 1)(x2 + 3x – 2x – 6)

= (x + 1){x(x + 3) – 2(x + 3)}

= (x + 1)(x + 3)(x – 2)

(ii) f(x) = x3 – 13x – 12

Let x = 4, then

f(x) = (4)3 – 13(4) – 12

= 64 – 52 – 12

= 64 – 64 = 0

∵ f(x) = 0

∴ x – 4 is a factor of f(x).

Now dividing f(x) by (x – 4), we get,

f(x) = (x – 4)(x2 + 4x + 3)

= (x – 4)(x2 + 3x + x + 3)

= (x – 4)[x(x + 3) + 1(x + 3)]

Question 16: Use the remainder Theorem to factorise the following expression

(i) 2x3 + x2 – 13x + 6.

(ii) 3x2 + 2x2 – 19x + 6

(iii) 2x3 + 3x2 – 9x – 10

(i) Let f(x) = 2x3 + x2 – 13x + 6

Factors of 6 are ±1, ±2, ±3, ±6

Let x = 2, then

f(2) = 2(2)3 + (2)2 – 13 × 2 + 6

= 16 + 4 – 26 + 6

= 26 – 26

= 0

∵  f(2) = 0

∴  x – 2 is the factor of f(x) (By Remainder Theorem)

Dividing f(x) by x – 2, we get

∴ f(x) = (x – 2)(2x2 + 5x – 3)

= (x – 2){2x2 + 6x – x – 3}

= (x – 2){2x(x + 3) – 1(x + 3)}

= (x – 2)(x + 3)(2x – 1)

(ii) P(x) = 3x3 + 2x2 – 19x + 6

P(1) = 3 + 2 – 19 + 6

= - 8 ≠ 0

P(-1)= - 3 + 2 + 19 + 6

= - 24 ≠ 0

P(2) = 24 + 8 – 38 + 6 = 0

Hence, (x – 2) is a factor of p(x)

∴ P(x) = 3x3 + 2x2 – 19x + 6

= 3x3 – 6x2 + 8x2 – 16x – 3x + 6

= 3x2(x – 2) + 8x(x – 2) – 3(x – 2)

= (x – 2)(3x2 + 8x – 3)

= (x – 2)(3x2 + 9x – x – 3)

= (x – 2){3x(x + 3) – 1(x + 3)

= (x – 2)(x + 3)(3x – 1)

(iii) f(x) = 2x3 + 3x2 – 9x - 10

f(-1) = 2(-1)3 + 3(-1)2 – 9(-1) – 10

∴ f(-1) = - 2 + 3 + 9 – 10 = 0

∴ (x + 1) is a factor.

∴ 2x2 + x – 10 = 2x2 + 5x – 4x – 10

= x(2x + 5) – 2(2x + 5) – (2x + 5)(x – 2)

∴ factors are (x + 1) (x – 2)(2x + 5)

17. Using the Remainder and factor Theorem, factorise the following polynomial:

x3 + 10x2 – 37x + 26.

f(x) = x3 + 10x2 – 37x + 26

f(1) = (1)3 + 10(1)2 – 37(1) + 26

= 1 + 10 – 37 + 26

= 0

x = 1

x – 1 is factor of f(x)

∴ f(x) = (x – 1)(x2 + 11x – 26)

= (x – 1)(x2 + 13x – 2x – 26)

= (x – 1)[x(x + 13) – 2(x + 13)]

= (x – 1)[(x – 2)(x + 13)]

18. If (2x + 1) is a factor of 6x3 + 5x2 + ax – 2. Find the value of a.

Let 2x + 1 = 0, then x = - (1/2)

Substituting the value of x in f(x),

f(x) = 6x3 + 5x2 + ax – 2

f(-1/2) = 6(- 1/2)3 + 5(- 1/2)2  + a(-1/2) – 2

= 6(-1/8) + 5(1/4) + a(- 1/2) – 2

= - 3/4 + 5/4 – a/2 – 2

= (-3 + 5 – 2a – 8)/4

= (- 6 – 2a)/4

∵ 2x + 1 is a factor of f(x)

∴ Remainder = 0

∴ (- 6 – 2a)/4 = 0

⇒ - 6 – 2a = 0

⇒ 2a = - 6

⇒ a = - 3

∴ a = - 3

19. If (3x – 2) is a factor of 3x3 – kx2 + 21x – 10, find the value of k.

Let 3x – 2 = 0, then 3x = 2

⇒ x = 2/3

Substituting the value of x in f(x),

f(x) = 3x3 – kx2 + 21x – 10

f(2/3) = 3(2/3)3 – k(2/3)2 + 21(2/3) – 10

= 3 × 8/27 – k × 4/9 + 21 × 2/3 – 10

= 8/9 – 4k/9 + 14 – 10

= (8 – 4k)/9 + 4

∵ Remainder is 0

∴ (8 – 4k)/9 + 4 = 0

⇒ 8 – 4k + 36 = 0

⇒ - 4k + 44 = 0

⇒ 4k = 44

∴ k = 11

20. If ( x – 2) is a factor of 2x3 –x2 + px – 2, then

(i) find the value of p.

(ii) with this value of p, factorize the above expression completely

(i) Let x – 2 = 0, then x = 2

Now f(x) = 2x3 – x2 + px – 2

∴ f(2) = 2(3)3 – (2)2 + p × 2 – 2

= 2 × 8 – 4 + 2p – 2

= 16 – 4 + 2p – 2 = 10 + 2p

(ii) ∴ f(2) = 0, then 10 + 2p = 0

⇒ 2p = -10

⇒ p = -5

Now, the polynomial will be

2x3 – x2 – 5x – 2

= (x – 2)(2x2 + 3x + 1)

= (x – 2)[2x2 + 2x + x + 1]

= (x – 2)[2x(x + 1) + 1(x + 1)]

= (x – 2)(x + 1)(2x + 1)

21. Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, (k + 2)x2 – Kx + 6 = 0. Also, find the other root of the equation.

(K + 2)x2 – Kx + 6 = 0 …(1)

Substitute x = 3 in equation (1)

(- 4 + 2)x2 – (- 4)x + 6 = 0

⇒ - 2x2 + 4x + 6 = 0

⇒ x2 – 2x – 3 = 0 (Dividing by 2)

⇒ x3 – 3x + x – 3 = 0

⇒ x2 – 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x + 1)(x – 3) = 0

So, the roots are x = - 1and x = 3

Thus, the other root of the equation is x = - 1

22. What number should be subtracted from 2x3 – 5x2 + 5x so that the resulting polynomial has 2x – 3 as a factor?

Let the number to be subtracted be k and the resulting polynomial be f(x), then

f(x) = 2x3 – 5x2 + 5x – k

Since, 2x – 3 is a factor of f(x),

Now, converting 2x – 3 to factor theorem

f(3/2) = 0

⇒ 2x3 – 5x2 + 5x – k = 0

⇒ 2(3/2)3 – 5(3/2)2 + 5(3/2) – k = 0

⇒ 2 × 27/8 – 5 × 9/4 + 5 × 3/2 – k = 0

⇒ 27/4 – 45/4 + 15/2 – k = 0

⇒ 27 – 45 + 30 – 4k = 0

⇒ - 4k + 12 = 0

⇒ k = - 12/-4

⇒ k = 3

23. (i) Find the value of the constant a and b, if (x – 2) and (x + 3) are both factors of the expression x3 + ax2 + bx – 12.

(ii) If (x + 2) and (x + 3) are factors of x3 + ax + b, Find the values of a and b.

Let x – 2 = 0, then x = 0

Substituting value of x in f(x)

f(x) = x3 + ax2 + bx – 12

f(2) = (2)3 + a(2)2 + b(2) – 12

= 8 + 4a + 2b – 12

= 4a + 2b – 4

∵ x – 2 is a factor

∴ 4a + 2b – 4 = 0

⇒ 4a + 2b = 4

⇒ 2a + b = 2

Again let x + 3 = 0, then x = - 3

Substituting the value of x in f(x)

f(x) = x3 + ax2 + bx – 12

= (-3)3 + a(-3)2 + b(-3) – 12

⇒ - 27 + 9x – 3b – 12

= - 39 + 9a – 3b

∵ x + 3 is a factor of f(x)

∴ - 39 + 9a – 3b = 0

⇒ 9a – 3b = 39

⇒ 3a – b = 13 ….(ii)

5a = 15

⇒ a = 3

Substituting the value of a in (i)

2(3) + b = 2

⇒ 6 + b = 2

⇒ b = 2 – 6

∴ b = - 4

Hence a = 3, b = - 4

(ii) Since (x + 2) = 0

f(x) = x3 + ax + b

f(-2) = (-2)3 + a(-2) + b

Since x + 2 is a factor, by factor theorem

- 8 – 2a + b = 0

∴ - 2a + b = 8 …..(i)

Since x + 3 = 0

∴ x = - 3

f(x) = x3 + ax + b

∴ f(-3) = (-3)3 + a(-3) + b

∴ f(-3) = - 27 – 3a + b

By factor theorem, - 27 – 3a + b = 0

∴ - 3a + b = 27 ….(ii)

Subtracting eq. (ii) from eq. (i)

- 2a + b = 8

- 3a + b = 27

a = - 19

Now, substituting the value of a in eq. (i)

- 2(- 19) + b = 8

⇒ 38 + b = 8

b = - 30

24. If (x + 2) and (x – 3) are factors of x3 + ax + b, find the values of a and b, With these values of a and b, factorize the given expression.

Let x + 2 = 0, then x = - 2

Substituting the value of x in f(x),

f(x) = x3 + ax + b

f(-2) = (-2)3 + a(-2) + b

= - 8 – 2a + b

∵ x + 2 is a factor

∴ Remainder is zero.

∴ - 8 – 2a + b = 0

⇒ - 2a + b = 8

∴ 2a – b = - 8 ...(i)

Again let x – 3 = 0, then x = 3, Substituting the value of x in f(x),

f(x) = x3 + ax +b

f(3) = (3)3 + a(3) + b

= 27 + 3a + b

∵ x – 3 is a factor

∴ Remainder = 0

⇒ 27 + 3a + b = 0

⇒ 3a + b = - 27 ...(ii)

5a = - 35

⇒ a = - 35/5

= - 7

Substituting value of a in (i)

2(-7) – b = - 8

∴ - 14 – b = - 8

- b = - 8 + 14

⇒ - b = 6

∴ b = - 6

Hence a= - 7, b = - 6

(x + 2) and (x – 3) are the factors of x2 + ax + b

x3 + ax + b ⇒ x3 + 7x - 6

Now dividing x3 – 7x – 6

Now dividing x3 – 7x – 6 by (x + 2)

(x – 3) or x2 – x – 6, we get

∴ Factors are (x + 2), (x – 3) and (x + 1)

25. (x – 2) is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b.

As, x – 2 is a factor of

f(x) = x3 + ax2 + bx + 6

∴ f(2) = 0

∴ (2)3 + a(2)2 + b(2) + 6 = 0

⇒ 8 + 4a + 2b + 6 = 0

⇒ 4a + 2b = - 14

⇒ 2a + b = - 7 ...(i)

As on dividing f(x) by x – 3

Remainder = 3

∴ f(3) = 3

∴ (3)3 + a(3)2 + b(3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b = - 30

⇒ 3a + b = - 10 ...(ii)

Solving simultaneously equation (i) and (ii),

∴ 2a + b = - 7

3a + b = - 10

Subtracting,

- a = 3

a = - 3

Substituting value of a in equation (i)

2(-3) + b = - 7

∴ 6 + b = - 7

∴ b = - 1

∴ a = - 3, b = - 1

26. If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divide by (x – 3), it leaves a remainder 52, find the values of a and b.

f(x) = 2x3 + ax2 +bx – 14

∴ (x – 2) is factor of f(x)

f(2) = 0

2(2)3 + a(2)2 + b(2) – 14 = 0

16 + 4a + 2b – 14 = 0

⇒ 4a + 2b = - 2

2a + b = - 1 ….(i)

Also, (x – 3) it leaves remainder = 52

∴ f(3) = 52

2(3)3 + a(3)2 + b(3) – 14 = 52

⇒ 54 + 9a + 3b – 14 = 52

⇒ 9a + 3b = 52 – 40

9a + 3b = 12

3a + b = 4 ...(ii)

From (i) and (ii)

2a + b = - 1

3a + b = 4

Subtracting – a = - 5

∴ a = 5 put in (i)

∴ 2(5) + b = - 1

⇒ b = - 1 – 10

⇒ b = - 11

∴ a = 5, b = - 11

⇒ - 27a/8 + 27/4 – 3b/2 – 3 = 0

⇒ - 27a + 54 – 12b – 24 = 0 (Multiplying by 8)

⇒ - 27a – 12b + 30 = 0

⇒ - 27 a – 12b = - 30

⇒ 9a + 4b = 10 [Dividing by (-3)]

9a + 4b = 10 …..(i)

Again let x + 2 = 0 then x = - 2

Substituting the value of x in f(x)

f(x) = ax3 + 3x2 + bx – 3

f(-2) = a(-2)3 + 3(-2)2 + b(-2) – 3

= - 8a + 12 – 2b – 3

= - 8a – 2b + 9

∵ Remainder = - 3

∴ - 8a – 2b + 9 = - 3

⇒ - 8a – 2b = - 3 – 9

⇒ - 8a -2b = - 12 (Dividing by 2)

⇒ 4a + b = 6 …(ii)

Multiplying (ii) by 4

16a + 4b = 24

9a + 4b = 10

Subtracting, 16a + 4b = 24

7a = 14

7a = 14

⇒ a = 14/7 = 2

Subtracting the value of a in (i)

9(2) + 4b = 10

⇒ 18 + 4b = 10

⇒ 4b = 10 – 18

⇒ 4b = - 8

∴ b = - 8/4 = - 2

Hence a = 2, b = - 2

∴ f(x) = ax3 + 3x2 + bx – 3

= 2x3 + 3x2 – 2x – 3

∵ 2x + 3 is a factor

∴ Dividing f(x) by x + 2

∴ 2x3 + 3x2 – 2x – 3

= (2x + 3)(x2 – 1) = (2x + 3)[(x)2 – (1)2]

= (2x + 3)(x + 1)(x – 1)

27. If ax3 + 3x2 + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and 6. With these values of a and 6, factorise the given expression.

Let 2x + 3 = 0 then 2x = - 3

⇒ x = (-3)/2

Substituting the value of x in f(x),

f(x) = ax3 + 3x2 + 6x – 3

f(-3/2) = a(-3/3)3 + 3(-3/2)2 + b(-3/2) – 3

= a(- 27/8) + 3(9/4) + b(-3/2) - 3

= - 27a/8 + 27/4 – 3b/2 – 3

∵ 2x + 3 is a factor of f(x)

∴ Remainder = 0

28. Given f(x) = ax2 + bx + 2 and g(x) = bx2 + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x) + g(x) + 4x2 + 7x.

f(x) = ax2 + bx + 2

g(x) = bx2 + ax + 1

x – 2 is a factor of f(x)

Let x – 2 = 0

⇒ x = 2

∴ f(2) = a(2)2 + b × 2 + 2

= 4a + 2b + 2

∴ 4a + 2b + 2 = 0 (∵ x – 2 is its factor)

⇒ 2a + b + 1 = 0 ….(i) (Dividing by 2)

Dividing g(x) by x – 2, remainder = - 15

Let x – 2 = 0

⇒ x = 2

∴ g(2) = b(2)2 + a × 2 + 1

= 4b + 2a + 1

∵ Remainder is – 15

∴ 4b + 2a + 1 = - 15

⇒ 4b + 2a + 1 + 15 = 0

⇒ 4b + 2a + 16 = 0

⇒ 2b + a + 8 = 0 (Dividing by 2)

⇒ a + 2b + 8 = 0 …(ii)

Multiplying (i) by 2 and (ii) by 1

4a + 2b + 2 = 0

a + 2b + 8 = 0

3a – 6 = 0

⇒ 3a = 6

⇒ a = 6/3

∴ a = 2

Substituting the value of a in (i)

2 × 2 + b + 1 = 0

⇒ 4 + b + 1 = 0

⇒ b + 5 = 0

⇒ b = - 5

Hence a = 2, b = - 5

Now f(x) + g(x) = 4x2 + 7x

= 2x2 – 5x + 2 + (-5x2 + 2x + 1) + 4x2 + 7x

= 2x2 – 5x + 2 – 5x2 + 2x + 1 + 4x2 + 7x

= 6x2 – 5x2 – 5x + 2x + 7x + 2 + 1

= x2+ 2x + 3

= x2 + x + 3x + 3

= x(x + 1) + 3(x + 1)

= (x + 1)(x + 3)

### Multiple Choice Questions

Choose the correct answer from the given four options (1to 5):

1. When x3 – 3x2 + 5x – 7 is divided by x – 2, then the remainder is. .

(a) 0

(b) 1

(c) 2

(d) – 1

(d) – 1

f(x) = x3 – 3x2 + 5x – 7

g(x) = x – 2, if x – 2 = 0, then x = 2

Remainder will be

∴ f(2) = (2)2 – 3(2)2 + 5 × 2 – 7

= 8 – 12 + 10 – 7

= 18 – 19

= - 1

∴ Remainder = - 1

2. When 2x3 – x2 – 3x + 5 is divided by 2x + 1, then the remainder is

(a) 6

(b) – 6

(c) – 3

(d) 0

(a) 6

f(x) = 2x3 – x2 – 3x + 5

g(x) = 2x + 1

Let 2x + 1 = 0, then x = -1/2

Then remainder will be

f(-1/2) = 2(-1/2)3 – (-1/2)2 – 3(-1/2) + 5

= 2 × -1/8 – 1/4 + 3/2 + 5

= -1/4 – 1/4 + 3/2 + 5

= (- 1 – 1 + 6 + 20)/4

= 24/4

= 6

∴ Remainder = 6

3. If on dividing 4x2 – 3kx + 5 by x + 2, the remainder is – 3 then the value of k is

(a) 4

(b) – 4

(c) 3

(d) – 3

(b) – 4

f(x) = 4x2 – 3kx + 5

g(x) = x + 2

Remainder = - 3

Let x + 2 = 0, then x = - 2

Now remainder will be

f(-2) = 4(-2)2 – 3k(-2) + 5

= 16 + 6k + 5

= 21 + 6k

∴ 21 + 6k = - 3

⇒ k = - 24/6

= - 4

∴ k = - 4

4. If on dividing 2x3 + 6x2 – (2k – 7)x + 5 by x + 3, the remainder is k – 1 then the value of k is

(a) 2

(b) – 2

(c) – 3

(d) 3

(d) 3

f(x) = 2x3 + 6x2 – (2k – 7)x + 5

g(x) = x + 3

Remainder = k – 1

If x + 3 = 0, then x = - 3

∴ Remainder will be

f(-3) = 2(-2)3 + 6(-3)2 – (2k – 7) (-3) + 5

= -54 + 54 + 3(2k – 7) + 5

= -54 + 54 + 6k – 21 + 5

= 6k – 16

∴ 6k – 16 = k – 1

6k – k = - 1 + 16

⇒ 5k = 15

⇒ k = 15/5 = 3

∴ k = 3

5. If x + 1 is a factor of 3x3 + kx2 + 7x + 4, then the value of k is

(a) – 1

(b) 0

(c) 6

(d) 10

(c) 6

f(x) = 3x3 + kx2 + 7x + 4

g(x) = x + 1

Remainder = 0

Let x + 1 = 0, then x = - 1

f(-1) = 3(-1)3 + k(-1)2 + 7(-1) + 4

= - 3 + k – 7 + 4

= k – 6

∴ Remainder = 0

∴ k – 6 = 0

⇒ k = 6

### Chapter Test

1. Find the remainder when 2x3 – 3x2 + 4x + 7 is divided by

(i) x – 2

(ii) x + 3

(iii) 2x + 1

f(x) = 2x3 – 3x2 + 4x + 7

(i) Let x – 2 = 0, then x = 2

Substituting value of x in f(x)

f(2) = 2(2)3 – 3(2)2  + 4(2) + 7

= 2 × 8 – 3 × 4 + 4 × 2 + 7

= 16 – 12 + 8 + 7

= 19

Remainder = 19

(ii) Let x + 3 = 0, then x = - 3

Substituting the value of x in f(x)

f(-3) = 2(-3)3 – 3(-3)2 + 4(-3) + 7

= 2 ×(-27) – 3(9) + 4(-3) + 7

= - 54 – 27 – 12 + 7

= - 93 + 7

= - 86

∴ Remainder = - 86

(ii) Let 2x + 1 = 0, then 2x = - 1

⇒ x = - (1/2)

Now substituting the value of x in f(x)

f(- 1/2) = 2(- 1/2)3 – 3(- 1/2)2 + 4(- 1/2) + 7

= 2 ( - 1/8) – 3(1/4) + 4(- 1/2) + 7

= - 1/4 – 3/4 – 2 + 7

= - 1 – 2 + 7

= 4

∴ Remainder = 4

2. When 2x3 – 9x2 + 10x – p is divided by (x + 1), the remainder is – 24. Find the value of p.

Let x + 1 = 0 then x = - 1

Substituting the value of x in f(x)

f(x) = 2x3 – 9x2 + 10x – p

f(-1) = 2(-1)3 – 9(-1)2 + 10(-1) – p

= - 2 – 9 – 10 – p

= - 21 – p

∵ - 21 – p = - 24

⇒ - p = - 24 + 21 = - 3

∴ p = 3

3. If (2x – 3) is a factor of 6x2 + x + a, find the value of a. With this value of a, factorise the given expression.

Let 2x – 3 = 0 then 2x = 3

⇒ x = 3/2

Substituting the value of x in f(x)

f(x) = 6x2 + x + a

f(3/2) = 6(3/2)2 + 3/2 + a

= 6 × 9/4 + 3/2 + a

= 27/2 + 3/2 + a

= 30/2 + a

= 15 + a

∴ 2x – 3 is the factor

∴ Remainder = 0

∴ 15 + a = 0

⇒ a = - 15

Now f(x) will be 6x2 + x – 15 s

Dividing 6x2 + x – 15 by 2x – 3, we get

∴ 6x2 + x – 15 = (2x – 3)(3x + 5)

4. When 3x2 – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also the polynomial 3x2 – 5x + p – 3.

f(x) = 3x2 – 5x + p

Let (x – 2) = 0, then x = 2

f(2) = 3(2)2 – 5(2) + p

= 3 × 4 – 10 + p,

= 12 – 10 + p

= 2 + p

∵ Remainder = 3

∴ 2 + p = 3

⇒ p = 3 – 2 = 1

Hence p = 1

Now f(x) = 3x2 – 5x + p – 3

= 3x2 – 5x + 1 – 3

= 3x2 – 5x – 2

Dividing by (x – 2), we get

3x2 – 5x – 2 = (x – 2)(3x + 1)

5. Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely.

f(x) = 5x3 + 4x2 – 5x – 4

Let 5x + 4 = 0, then 5x = - 4

⇒ x = - 4/2

∴ f( - 4/5) = 5(- 4/5)3 + 4(- 4/5)2 – 5( - 4/5) – 4

= 5 × ( - 64/125) + 4 × 16/25 + 4 – 4

= - 64/25 + 64/25 + 4 – 4

= 0

∵ f(-4/5) = 0

∴ (5x + 4) is a factor of f(x)

Now, dividing f(x) by 5x + 4, we get

5x3 + 4x2 – 5x – 4

= (5x + 4)(x2 – 1)

= (5x + 4){(x)2 – (1)2}

= (5x + 4)(x + 1)(x – 1)

6. Use factor theorem to factorise the following polynomials completely:

(i) 4x3 + 4x2 – 9x – 9

(ii) x – 19x – 30

(i) f(x) = 4x3 + 4x2 – 9x – 9

Let x = - 1, then

f(-1) = 4(1)3 + 4(-1)2 – 9(1) – 9

= - 4 + 4 + 9 – 9

= 13 – 13

= 0

∴ (x + 1) is a factor of f(x)

Now dividing f(x) by x + 1, we get

f(x) = 4x3 + 4x2 – 9x – 9

= (x + 1)(4x2 – 9)

= (x + 1){(2x)2 – (3)2}

= (x + 1)(2x + 3)(2x – 3)

(ii) f(x) = x3 – 19x – 30

Let x = - 2, then

f(-2) = (-2)2 – 19(-2) – 30

= - 8 + 38 – 30

= - 38 – 38

= 0

∴ (x + 2) is a factor of f(x).

Now dividing f(x) by (x + 2), we get

f(x) = x3 – 19x – 30

= (x + 2) (x2 – 2x – 15)

= (x + 2){(x2 – 5x + 3x – 15)}

= (x + 2){x(x – 5) + 3(x – 5)}

= (x + 2)(x – 5)(x + 3)

7. If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, find the values of p and q. With these values of p and q, factorize the given polynomial completely.

f(x) = x3 – 2x2 + px + q

(x + 2) is a factor

f(-2) = (-2)3 – 2(-2)2 + p(-2) + q

= - 8 – 2 × 4 – 2p + q

= - 8 – 8 – 2p + q

= - 16 – 2p + q

∵ (x + 2) is a factor of f(x)

∴ f(-2) = 0

⇒ - 16 – 2p + q = 0

⇒ 2p – q = - 16 ...(i)

Again, let x + 1 = 0, then x = - 1

∴ f(-1) = (- 1)3 – 2(-1)2 + p(-1) + q

= - 1 – 2 × 1 – p + q

= - 1 – 2 – p + q

= - 3 – p + q

∵ Remainder = 9, then

- 3 – p + q = 9

⇒ - p + q = 9 + 3 = 12

- p + q = 12 ...(ii)

p = - 4

Substituting the value of p in (ii)

- (- 4) + q = 12

4 + q = 12

⇒ q = 12 – 4 = 8

∴ p = - 4, q = 8

∴ f(x) = x3 – 2x2 – 4x + 8

Dividing f(x) by (x + 2), we get

f(x) = (x + 2)(x2 – 4x + 4)

= (x + 2){(x)2 – 2 × x(-2) + (2)2}

= (x + 2)(x – 2)2

8. If (x + 3) and (x – 4) are factors of x3 + ax2 – bx + 24, find the values of a and b: with these values of a and b, factorise the given expression.

f(x) = x3 + ax2 – bx + 24

Let x + 3 = 0, then x = - 3

Substituting the value of x in f(x)

f(-3) = (-3)3 + a(- 3)2 – b(-3) + 24,

= - 27 + 9a + 3b + 24

= 9a + 3b – 3

∵ Remainder = 0,

∴ 9a + 3b – 3 = 0

⇒ 3a + b – 1 = 0 (Dividing by 3)

⇒ 3a + b = 1 ...(i)

Again Let x – 4 = 0, then x = 4

Substituting the value of x in f(x)

f(x) = (4)3 + a(4)2 – b(4) + 24

= 64 + 16a- 4b + 24

= 16a – 4b + 88

∵ x – 4 is a factor

∴ Remainder = 0

16a – 4b + 88 = 0

⇒ 16a – 4b = - 88 (Dividing by 4)

⇒ 4a – b = - 22

7a = - 21,

⇒ a = - 3

Substituting the value of a in (i)

3(-3) + b = 1

⇒ - 9 + b = 1

⇒ b = 1 + 9 = 10

Now f(x) will be

f(x) = x3 – 3x2 – 10x + 24

∵ x + 3 and x – 4 are factors of f(x)

∴ Dividing f(x) by (x + 3)(x – 4)

or x2 – x – 12

x3 – 3x2 – 10x + 24

= (x2 – x – 12)(x – 2)

= (x + 3)(x – 4)(x – 2)

9. If 2x3 + ax2 – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.

f(x) = 2x3 + ax2 – 11x + b

Let x – 2 = 0, then x = 2,

Substituting the value of x in f(x)

f(2) = 2(2)3 + a(2)2 – 11(2) + b

= 2 × 8 + 4a – 22 + b

= 16 + 4a – 22 + b

= 4a + b – 6

∵ Remainder = 0,

∴ 4a + b – 6 = 0

⇒ 4a + b = 6 …(i)

Again let x – 3 = 0, then x = 3

Substituting the value of x is f(x)

f(3) = 2(3)3 + a(3)2 – 11 × 3 + b

= 2 × 27 + 9a – 33 + b

= 54 + 9a – 33 + b

⇒ 9a + b + 21

∵ Remainder = 42

∴ 9a + b + 21 = 42

⇒ 9a + b = 42 – 21

⇒ 9a + b = 21 ….(ii)

Subtracting (i) from (ii)

5a = 15

⇒ a = 15/5 = 3

Substituting the value of a is (i) 4(3) + b = 6

⇒ 12 + b = 6

⇒ b = 6 – 12

⇒ b = - 6

∴ f(x) will be 2x3 + 3x2 – 11x – 6

∵ x – 2 is a factor (as remainder = 0)

∴ Dividing f(x) by x – 2, we get

∴ 2x3 + 3x2 – 11x – 6

= (x – 2)(2x2 + 7x + 3)

= (x – 2)[2x2 + 6x + x + 3]

= (x – 2)[2x(x + 3) + 1(x + 3)]

= (x – 2)(x + 3)(2x + 1)

10. If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.

Let 2x + 1 = 0, then 2x = - 1

x = - (1/2)

Substituting the value of x in

f(x) = 2x2 – 5x + p

f(-1 /2) = 2(-1/2)2 – 5(-1/2) + p

= 2 × 1/4 + 5/2 + p

= 1/2 + 5/2 + p

= 3 + p

∵ 2x + 1 is the factor of p(x)

∴ Remainder = 0

⇒ 3 + p = 0

⇒ p = - 3

Again substituting the value of x in q(x)

q(x) = 2x2 + 5x + q

q(- 1/2) = 2(- 1/2)2 + 5(- 1/2) + q

= 2 × 1/4 – 5/2 + q

= 1/2 – 5/2 + q

= - 4/2 + q

= q – 2

∵ 2x + 1 is the factor of q(x)

∴ Remainder = 0

⇒ q – 2 = 0

⇒ q = 2

Hence p = - 3, q = 2

Now (i) ∵ 2x + 1 is the factor of p(x)

= 2x2 – 5x – 3

∴ Dividing p(x) by 2x + 1,

∴ 2x2 – 5x – 3 = (2x + 1)(x – 3)

(ii) ∵ 2x + 1 is the factor of q(x) = 2x2 + 5x + 2

∴ Dividing q(x) by 2x + 1,

∴ 2x2 + 5x + 2 = (2x + 1)(x + 2)

11. If a polynomial f(x) = x4 – 2x3 – 3x2 – ax – b leaves remainder 5 and 19 when divided by (x – 1) and (x + 1) respectively, Find the values of a and b. Hence determined the remainder when f(x) is divided by (x – 2).

f(x) = x4 – 2x3 + 3x2 – ax + b

f(1) = 5 and f(-1) = 19

∴ (1)4 – 2(1)3 + 3(1)2 – a (1) + b = 5

And (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b = 19

⇒ 1 – 2 + 3 – a + b = 5

and 1 + 2 + 3 + a + b = 19

⇒ - a + b = 5 – 2 and a + b = 19 – 6

⇒ a+ b = 3 ….(i)

and a + b = 13 ….(ii)

On subtracting (i) and (2), we het

a + b – (- a + b) = 13 – 3

a + b + a – b = 10

2a = 10

a = 5

putting a = 5 in equation 1, we get

= 5 + b = 3, b = 8

a = 5, b = 8

12. When a polynomial f(x) is divided by (x – 1), the remainder I 5 and when it is, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1)(x – 2).

When f(x) id divided by (x – 1),

Remainder = 5

Let x – 1 = 0

⇒ x = 1

∴ f(1) = 5

When divided by (x – 2), Remainder = 7

Let x – 2 = 0

⇒ x = 2

∴ f(2) = 7

Let f(x) = (x – 1)(x – 2)q(x) + ax + b

Where q(x) is the quotient and ax + b is remainder

Putting x = 1, we get:

f(1) = (1 – 1) (1 – 2) q(1) + a × 1 + b

= 0 + a + b

= a + b

And x = 2, then

f(2) = (2 – 1)(2 – 2) q(2) + a ×1 + b

= 0 + 2a + b

= 2a + b

∴ a + b = 5 …(i)

2a + b = 7 ….(ii)

Subtracting, we get

- a = - 2

⇒ a = 2

Substituting the value of a in (i)

2 + b = 5

⇒ b = 5 – 2 = 3

∴ a = 2, b = 3

∴ Remainder = ax + b

= 2x + 3

The solutions provided for Chapter 6 Factorization of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 6 Factorization contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.

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