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

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

**(i) f(x) = 5x ^{2} – 1x + 4 **

**(ii) f(x) = 2x ^{3} – 7x^{2} + 3**

**Answer**

Let x – 2 = 0, then x = 2

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

f(x) = 5x^{2} – 7x + 4

⇒ f(2) = 5(2)^{2} – 7(2) + 4

⇒ f(2) = 20 – 14 + 4 = 10

Hence Remainder = 10

**(ii)** f(x) = 2x^{3} – 7x^{2 }+ 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) = 2x ^{2} – 5x + 1 **

**(ii) f(x) = 3x ^{3} + 7x^{2 }– 5x + 1 **

**Answer**

Let x + 3 = 0

⇒ x = - 3

Substituting the value of x in f(x),

**(i)** f(x) = 2x^{2 }– 5x + 1

∴ f(-3) = 2(-3)^{2} – 5(-3) + 1

= 18 + 15 + 1

= 34

Hence Remainder = 34

**(ii)** f(x) = 3x^{3} + 7x^{2} – 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) = 4x ^{2} + 5x + 3**

**(ii) f(x) = 3x ^{2} – 7x^{2 }+ 4x + 11**

**Answer**

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

Substituting the value of x in f(x):

**(i)** f(x) = 4x^{2 }+ 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) = 3x^{3} – 7x^{2} + 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 2x ^{3} – 3x^{2 }+ 7x – 8 is divided by x – 1 **

**(ii) Find the remainder (without division) on dividing 3x ^{2 }+ 5x – 9 by (3x + 2) **

**Answer**

**(i)** Let x – 1 = 0, then x = 1

Substituting value of x in f(x)

f(x) = 2x^{3} – 3x^{2 }+ 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) = 3x^{2} + 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 2****x ^{2}**

**+ 3**

**x**

^{2}**– kx + 5 by x – 2, leaves a remainder 7.**

**Answer**

f(x) = 2x^{2} + 3x^{2} – 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 x ^{3} + 5x^{2} – ax + 6 by (x – 1) leaves the remainder 2a. **

**Answer**

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x)

f(x) = x^{3} + 5x^{2} – 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 2x ^{2} – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?**

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

**Answer**

**(i) **Let a be subtracted from 2x^{2 }– 5x,

Dividing 2x^{2} – 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 2x^{3} – 7x^{2 }+ 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 x ^{2} – px^{2} + x + 6 and 2x^{3} – x^{2} – (p + 3)x – 6 leave the same remainder. Find the value of ‘p’ **

**(ii) Find ‘a’ if the two polynomials ax ^{3} + 3x^{2} – 9 and 2x^{3} + 4x + a, leaves the same remainder when divided by x + 3. **

**Answer**

**(i) **By dividing x^{3} – px^{2} + x + 6

And 2x^{3 }– x^{2} – (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) = x^{3} – px^{2} + x + 6

p(3) = (3)^{3} – p(3)^{2} + 3 + 6

= 27 – 9p + 9

= 36 – 9p

and q(x) = 2x^{3} – x^{2} – (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 ax^{3} + 3x^{2} – 9 and 2x^{3} + 4x + a, leaves the same remainder when divided by x + 3.

The given polynomials are ax^{3} + 3x^{2} – 9 and 2x^{3} + 4x + a

Let p(x) = ax^{3} + 3x^{2 }– 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 2x**^{2} **+ 5x – 3. **

**Answer**

Let x + 3 = 0 then x = - 3

Substituting the value of x in f(x)

f(x) = 2x^{2 }+ 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) = 2x^{2 }+ 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 x ^{4} + 2x^{3 }– 2x^{2} + 2x – 3 is exactly divisible by x^{2 }+ 2x – 3. **

**Answer**

x^{2} + 2x – 3

= x^{2} + 3x – x – 3

= x(x + 3) – 1(x + 3)

= (x + 3)(x – 1)

Let p(x) = x^{4} + 2x^{3} – 2x^{2} + 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., x^{2} + 2x – 3 is a factor of p(x)

⇒ p(x) is exactly divisible by (x^{2} + 2x – 3)

**11. Show that (x – 2) is a factor of 3x ^{2} – x – 10. Hence factorise 3x^{2 }– x – 10. **

**Answer**

Let x – 2 = 0, then x = 2

Substituting the value of x in f(x),

f(x) = 3x^{2} – x – 10

= 3(2)^{2} – 2 – 10

= 12 – 2 – 10 = 0

∵ Remainder is zero

∴ x – 2 is a factor of f(x).

Dividing 3x^{2} – x – 10 by x – 2, we get

∴ 3x^{2} – x – 10 = (x – 2)(3x + 5)

**12. (i) Show that (x – 1) is a factor of x ^{3} – 5x^{2} – x + 5. Hence, factorise x^{3} – 5x^{2 }– x + 5. **

**(ii) Show that (x – 3) is a factor of x ^{3 }– 7x^{2} + 15x – 9. Hence, factorise x^{3 }– 7x^{2} + 15x – 9 **

**Answer**

**(i) **Let x – 1 = 0, then x = 1

Substituting the value of x in f(x),

f(x) = x^{3} – 5x^{2} – x + 5

= (1)^{3} – 5(1)^{2} – 1 + 5

= 1 – 5 – 1 + 5

= 0

∵ Remainder = 0

∴ x – 1 is a factor of x^{3} – 5x^{2} – x + 5

Now dividing f(x) by x – 1, we get

∴ x^{3} – 4x^{2} – x + 5

= (x – 1)(x^{2} – 4x – 5) = (x – 1)[x^{2} – 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 x^{3} – 7x^{2} + 15x – 9

Now dividing it by x – 3, we get

∴ x^{3} – 7x^{2} + 15x – 9

= (x – 3)(x^{2} – 4x + 3) = (x – 3)[x^{2} – 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 4x**^{3} ** + 12x**^{2} ** + 11x + 3. Hence factorise 4**x^{3} **+ 12**x^{3} ** + 11x + 3.**

**Answer**

Let 2x + 1 = 0,

Then x = - 1/2

Substituting the value of x in f(x),

f(x) = 4x^{3} + 12x^{2} + 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

4x^{3} + 12x^{2} + 11x + 3

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

∴ 4x^{3 }+ 12x^{2} + 11x + 3

= (2x + 1)(2x^{2} + 5x + 3)

= (2x + 1)[2x^{2 }+ 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 2x**^{3}** + 5x**^{2}** – 11x – 14. Hence factorise the given expression completely, using the factor theorem.**

**Answer**

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

x = (-7)/2

Substituting the value of x in f(x),

f(x) = 2x^{3} + 5x^{2 }– 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, 2x^{3} + 5x^{2} – 11x – 14 = (2x + 7)(x^{2 }– x – 2)

= (2x + 7)[x^{2} – 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) x ^{3} + 2x^{2 }– 5x – 6 **

**(ii) x ^{3 }– 13x – 12**

**Answer**

**(i) **Let f(x) = x^{3 }+ 2x^{2} – 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) (x^{2} + x – 6)

= (x + 1)(x^{2} + 3x – 2x – 6)

= (x + 1){x(x + 3) – 2(x + 3)}

= (x + 1)(x + 3)(x – 2)

**(ii)** f(x) = x^{3} – 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)(x^{2} + 4x + 3)

= (x – 4)(x^{2} + 3x + x + 3)

= (x – 4)[x(x + 3) + 1(x + 3)]

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

**(i) 2x ^{3} + x^{2 }– 13x + 6. **

**(ii) 3x ^{2} + 2x^{2} – 19x + 6 **

**(iii) 2x ^{3} + 3x^{2 }– 9x – 10 **

**Answer**

**(i) **Let f(x) = 2x^{3} + x^{2 }– 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)(2x^{2} + 5x – 3)

= (x – 2){2x^{2} + 6x – x – 3}

= (x – 2){2x(x + 3) – 1(x + 3)}

= (x – 2)(x + 3)(2x – 1)

**(ii)** P(x) = 3x^{3} + 2x^{2} – 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) = 3x^{3 }+ 2x^{2} – 19x + 6

= 3x^{3} – 6x^{2} + 8x^{2 }– 16x – 3x + 6

= 3x^{2}(x – 2) + 8x(x – 2) – 3(x – 2)

= (x – 2)(3x^{2} + 8x – 3)

= (x – 2)(3x^{2 }+ 9x – x – 3)

= (x – 2){3x(x + 3) – 1(x + 3)

= (x – 2)(x + 3)(3x – 1)

(iii) f(x) = 2x^{3} + 3x^{2} – 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.

∴ 2x^{2} + x – 10 = 2x^{2} + 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:**

**x ^{3 }+ 10x^{2 }– 37x + 26. **

**Answer**

f(x) = x^{3} + 10x^{2} – 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)(x^{2} + 11x – 26)

= (x – 1)(x^{2} + 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 6x ^{3} + 5x^{2} + ax – 2. Find the value of a. **

**Answer**

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

Substituting the value of x in f(x),

f(x) = 6x^{3 }+ 5x^{2} + 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 3x ^{3 }– kx^{2 }+ 21x – 10, find the value of k. **

**Answer**

Let 3x – 2 = 0, then 3x = 2

⇒ x = 2/3

Substituting the value of x in f(x),

f(x) = 3x^{3} – kx^{2 }+ 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 2x ^{3} –x^{2} + px – 2, then **

**(i) find the value of p. **

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

**Answer**

**(i)** Let x – 2 = 0, then x = 2

Now f(x) = 2x^{3} – x^{2} + 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

2x^{3} – x^{2} – 5x – 2

= (x – 2)(2x^{2} + 3x + 1)

= (x – 2)[2x^{2} + 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)x ^{2} – Kx + 6 = 0. **

**Also, find the other root of the equation.**

**Answer**

(K + 2)x^{2} – Kx + 6 = 0 **…(1)**

Substitute x = 3 in equation (1)

(- 4 + 2)x^{2} – (- 4)x + 6 = 0

⇒ - 2x^{2 }+ 4x + 6 = 0

⇒ x^{2} – 2x – 3 = 0 (Dividing by 2)

⇒ x^{3} – 3x + x – 3 = 0

⇒ x^{2} – 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 2x ^{3} – 5x^{2} + 5x so that the resulting polynomial has 2x – 3 as a factor? **

**Answer**

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

f(x) = 2x^{3} – 5x^{2} + 5x – k

Since, 2x – 3 is a factor of f(x),

Now, converting 2x – 3 to factor theorem

f(3/2) = 0

⇒ 2x^{3} – 5x^{2} + 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 x ^{3} + ax^{2} + bx – 12. **

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

**Answer**

Let x – 2 = 0, then x = 0

Substituting value of x in f(x)

f(x) = x^{3} + ax^{2} + 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) = x^{3} + ax^{2 }+ 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)**

Adding (i) and (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) = x^{3 }+ 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) = x^{3} + 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 x ^{3 }+ ax + b, find the values of a and b, With these values of a and b, factorize the given expression. **

**Answer**

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

Substituting the value of x in f(x),

f(x) = x^{3} + 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) = x^{3 }+ 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)**

Adding (i) and (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 x^{2 }+ ax + b

x^{3} + ax + b ⇒ x^{3} + 7x - 6

Now dividing x^{3} – 7x – 6

Now dividing x^{3} – 7x – 6 by (x + 2)

(x – 3) or x^{2} – x – 6, we get

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

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

**Answer**

As, x – 2 is a factor of

f(x) = x^{3 }+ ax^{2 }+ 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 2x ^{3 }+ ax^{2 }+ bx – 14 and when the expression is divide by (x – 3), it leaves a remainder 52, find the values of a and b.**

**Answer**

f(x) = 2x^{3} + ax^{2} +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) = ax^{3} + 3x^{2} + 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

∴ f(x) = ax^{3} + 3x^{2} + bx – 3

= 2x^{3} + 3x^{2} – 2x – 3

∵ 2x + 3 is a factor

∴ Dividing f(x) by x + 2

∴ 2x^{3} + 3x^{2} – 2x – 3

= (2x + 3)(x^{2} – 1) = (2x + 3)[(x)^{2} – (1)^{2}]

= (2x + 3)(x + 1)(x – 1)

**27. If ax ^{3} + 3x^{2} + 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. **

**Answer**

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

⇒ x = (-3)/2

Substituting the value of x in f(x),

f(x) = ax^{3} + 3x^{2} + 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) = ax ^{2 }+ bx + 2 and g(x) = bx^{2} + 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) + 4x^{2} + 7x. **

**Answer**

f(x) = ax^{2} + bx + 2

g(x) = bx^{2 }+ 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) = 4x^{2} + 7x

= 2x^{2} – 5x + 2 + (-5x^{2 }+ 2x + 1) + 4x^{2} + 7x

= 2x^{2} – 5x + 2 – 5x^{2} + 2x + 1 + 4x^{2} + 7x

= 6x^{2} – 5x^{2} – 5x + 2x + 7x + 2 + 1

= x^{2}+ 2x + 3

= x^{2} + 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 x ^{3} – 3x^{2} + 5x – 7 is divided by x – 2, then the remainder is. .**

**(a) 0**

**(b) 1**

**(c) 2**

**(d) – 1 **

**Answer**

(d) – 1

f(x) = x^{3} – 3x^{2 }+ 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 2x ^{3 }– x^{2} – 3x + 5 is divided by 2x + 1, then the remainder is **

**(a) 6 **

**(b) – 6 **

**(c) – 3 **

**(d) 0 **

**Answer**

(a) 6

f(x) = 2x^{3 }– x^{2} – 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 4x ^{2 }– 3kx + 5 by x + 2, the remainder is – 3 then the value of k is **

**(a) 4 **

**(b) – 4 **

**(c) 3 **

**(d) – 3 **

**Answer**

(b) – 4

f(x) = 4x^{2} – 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 2****x ^{3} + 6x^{2}**

**– (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 **

**Answer**

(d) 3

f(x) = 2x^{3 }+ 6x^{2} – (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 3x ^{3} + kx^{2} + 7x + 4, then the value of k is **

**(a) – 1**

**(b) 0 **

**(c) 6 **

**(d) 10 **

**Answer**

(c) 6

f(x) = 3x^{3} + kx^{2} + 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 2x ^{3} – 3x^{2} + 4x + 7 is divided by **

**(i) x – 2 **

**(ii) x + 3 **

**(iii) 2x + 1 **

**Answer**

f(x) = 2x^{3} – 3x^{2} + 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 2x ^{3 }– 9x^{2} + 10x – p is divided by (x + 1), the remainder is – 24. Find the value of p. **

**Answer**

Let x + 1 = 0 then x = - 1

Substituting the value of x in f(x)

f(x) = 2x^{3} – 9x^{2 }+ 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 6x ^{2} + x + a, find the value of a. With this value of a, factorise the given expression.**

**Answer**

Let 2x – 3 = 0 then 2x = 3

⇒ x = 3/2

Substituting the value of x in f(x)

f(x) = 6x^{2} + 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 6x^{2} + x – 15 s

Dividing 6x^{2 }+ x – 15 by 2x – 3, we get

∴ 6x^{2} + x – 15 = (2x – 3)(3x + 5)

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

**Answer**

f(x) = 3x^{2} – 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) = 3x^{2} – 5x + p – 3

= 3x^{2} – 5x + 1 – 3

= 3x^{2} – 5x – 2

Dividing by (x – 2), we get

3x^{2} – 5x – 2 = (x – 2)(3x + 1)

**5. Prove that (5x + 4) is a factor of 5x ^{3} + 4x^{2} – 5x – 4. Hence factorize the given polynomial completely. **

**Answer**

f(x) = 5x^{3} + 4x^{2 }– 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

5x^{3} + 4x^{2} – 5x – 4

= (5x + 4)(x^{2} – 1)

= (5x + 4){(x)^{2} – (1)^{2}}

= (5x + 4)(x + 1)(x – 1)

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

**(i) 4x ^{3 }+ 4x^{2} – 9x – 9 **

**(ii) ****x ^{3 }**

**– 19x – 30**

**Answer**

**(i)** f(x) = 4x^{3} + 4x^{2} – 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) = 4x^{3} + 4x^{2} – 9x – 9

= (x + 1)(4x^{2} – 9)

= (x + 1){(2x)^{2} – (3)^{2}}

= (x + 1)(2x + 3)(2x – 3)

**(ii) **f(x) = x^{3 }– 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) = x^{3} – 19x – 30

= (x + 2) (x^{2} – 2x – 15)

= (x + 2){(x^{2} – 5x + 3x – 15)}

= (x + 2){x(x – 5) + 3(x – 5)}

= (x + 2)(x – 5)(x + 3)

**7. If x ^{3} – 2x^{2} + 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.**

**Answer**

f(x) = x^{3} – 2x^{2} + 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)**

Adding (i) and (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) = x^{3} – 2x^{2} – 4x + 8

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

f(x) = (x + 2)(x^{2} – 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 x ^{3} + ax^{2 }– bx + 24, find the values of a and b: with these values of a and b, factorise the given expression.**

**Answer**

f(x) = x^{3} + ax^{2} – 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

Adding (i) and (ii),

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) = x^{3 }– 3x^{2} – 10x + 24

∵ x + 3 and x – 4 are factors of f(x)

∴ Dividing f(x) by (x + 3)(x – 4)

or x^{2 }– x – 12

x^{3} – 3x^{2} – 10x + 24

= (x^{2} – x – 12)(x – 2)

= (x + 3)(x – 4)(x – 2)

**9. If 2x ^{3} + ax^{2} – 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.**

**Answer**

f(x) = 2x^{3 }+ ax^{2} – 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 2x^{3} + 3x^{2} – 11x – 6

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

∴ Dividing f(x) by x – 2, we get

∴ 2x^{3} + 3x^{2} – 11x – 6

= (x – 2)(2x^{2} + 7x + 3)

= (x – 2)[2x^{2} + 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 2x ^{2} – 5x + p and 2x^{2} + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.**

**Answer**

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

x = - (1/2)

Substituting the value of x in

f(x) = 2x^{2} – 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) = 2x^{2} + 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)

= 2x^{2} – 5x – 3

∴ Dividing p(x) by 2x + 1,

∴ 2x^{2 }– 5x – 3 = (2x + 1)(x – 3)

(ii) ∵ 2x + 1 is the factor of q(x) = 2x^{2 }+ 5x + 2

∴ Dividing q(x) by 2x + 1,

∴ 2x^{2} + 5x + 2 = (2x + 1)(x + 2)

**11. If a polynomial f(x) = x ^{4} – 2x^{3} – 3x^{2 }– 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). **

**Answer**

f(x) = x^{4} – 2x^{3} + 3x^{2} – 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). **

**Answer**

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

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