ML Aggarwal Factorisation Exe-6 Class 10 ICSE Maths Solutions Ch-6. We Provide Step by Step Answer of Exe-6 Questions for Factorisation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

### ML Aggarwal Factorisation Exe-6 Class 10 ICSE Maths Solutions Ch-6

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-6 | Factorisation |

Writer / Book | Understanding |

Topics | Solutions of Exe-6 |

Academic Session | 2024-2025 |

### ML Aggarwal Ch-6 **Factorisation Exe-6 **

**Class 10 ICSE Maths Solutions**

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

(i) f(x) = 5x^{2} – 7x + 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

= (5 × 4) – 14 + 4

= 20 – 14 + 4

= 24 – 14

= 10

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question- 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 :**

**(i) f(x) = 2x**^{2} – 5x + 1

^{2}– 5x + 1

Let x + 3 = 0

⇒ x = -3

Substituting the value of x in f(x),

f(2)= (2 × 2^{3}) – (7 × 2^{2}) + 3

= (2 × 8) – (7 × 4) + 3

= 16 – 28 + 3

= 19 – 28

= -9

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

^{3}+ 7x

^{2}– 5x + 1

Let x + 3 = 0

Then, x = -3

Given, f(x) = 3x^{3} + 7x^{2} – 5x + 1

Now, substitute the value of x in f(x),

f(-3)= (3 × -3^{3}) + (7 × -3^{2}) – (5 × -3) + 1

= (3 × -27) + (7 × 9) – (-15) + 1

= – 81 + 63 + 15 + 1

= -81 + 79

= -2

**Question -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^{3} – 7x^{2} + 4x + 11

**Answer:**

**(i) f(x) = 4x**^{2} + 5x + 3

^{2}+ 5x + 3

Let 2x + 1 = 0

Then, 2x = -1

X = -½

Given, f(x) = 4x^{2} + 5x + 3

Now, substitute the value of x in f(x),

f (-½) = 4 (-½)^{2} + 5 (-½) + 3

= (4 × ¼) + (-5/2) + 3

= 1 – 5/2 + 3

= 4 – 5/2

= (8 – 5)/2

= 3/2 = 1½

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

^{3}– 7x

^{2}+ 4x + 11

Let 2x + 1 = 0

Then, 2x = -1

X = -½

Given, f(x) = 3x^{3} – 7x^{2} + 4x + 11

Now, substitute the value of x in f(x),

f(-½) = (3 × (-½)^{3}) – (7 × (-½)^{2} + (4 × -½) + 11

= 3 × (-1/8) – (7 × ¼) + (- 2) + 11

= -3/8 – 7/4 – 2 + 11

= – 3/8 – 7/4 + 9

= (-3 – 14 + 72)/8

= 55/8

= 6(7/8)

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question-4. ****Using remainder theorem, find the value of k if on dividing 2x**^{3} + 3x^{2} – kx + 5 by x – 2, leaves a remainder 7.

^{3}+ 3x

^{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

Now, substitute the value of x in f(x),

f(2) = (2 × 2^{3}) + (3 × 2^{2}) – (k × 2) + 5

= (2 × 8) + (3 × 4) – 2k + 5

= 16 + 12 – 2k + 5

= 33 – 2k

given that, remainder = 7.

So, 7 = 33 – 2k

2k = 33 – 7

2k = 26

K = 26/2

K = 13

**Question -5. ****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.

^{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(1) = 1^{3}_{ }+ (5 × 1^{2}) – (a × 1) + 6

= 1 + 5 – a + 6

= 12 – a

given that, remainder = 2a

So, 2a = 12 – a

2a + a = 12

3a = 12

a = 12/3

a = 4

**Question -6**

**(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,

given that, remainder = 2.

3 – p = 2

p = 3 – 2

p = 1

**(ii) **let ‘p’ be subtracted from 2x^{3} – 7x^{2} + 2x,

given that, remainder = – 2.

P – 6 = – 2

P = -2 + 6

P = 4

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question -7**

**(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.**(iii) the polynomial ax**^{3}** + 3x**^{2}** – 3 and 2x**^{3}** -5x + a, when divide by x-4 leave the reminder r1 and r2 respectively. if 2r1 = r2 find the value of a.**

**Answer:**

**(i) given that, by dividing x**^{3} – px^{2} + x + 6 and 2x^{3} – x^{2} – (p + 3)x – 6 by x – 3 = 0,

^{3}– px

^{2}+ x + 6 and 2x

^{3}– x

^{2}– (p + 3)x – 6 by x – 3 = 0,

then x = 3.

Let p(x) = x^{3} – px^{2} + x + 6

Now, substitute the value of x in p(x),

p(3) = 3^{3} – (p × 3^{2}) + 3 + 6

= 27 – 9p + 9

= 36 – 9p

Then, q(x) = 2x^{3} – x^{2} – (p + 3)x – 6

Now, substitute the value of x in q(x),

q(3) = (2 × 3^{3}) – (3)^{2} – (p + 3) × 3 – 6

= (2 × 27) – 9 – 3p – 9 – 6

= 54 – 24 – 3p

= 30 – 3p

Given, that remainder is same in both case

So, 36 – 9p = 30 – 3p

36 – 30 = 9p – 3p

6 = 6p

p = 6/6

p = 1

**(ii) Let p(x) = ax**^{3} + 3x^{2} – 9 and q(x) = 2x^{3} + 4x + a

^{3}+ 3x

^{2}– 9 and q(x) = 2x

^{3}+ 4x + a

given that, both p(x) and q(x) leaves the same remainder when divided by x + 3.

Let x + 3 = 0

Then, x = -3

Now, substitute the value of x in p(x) and in q(x),

p(-3) = q(-3)

a(-3)^{3} + 3(-3)^{2} – 9 = 2(-3)^{3} + 4(-3) + a

-27a + 27 – 9 = – 54 – 12 + a

-27a + 18 = – 66 + a

-27a – a = -66 – 18

-28 a = -84

a = 84/28

a = 3

**(iii)** **the polynomial ax**^{3}** + 3x**^{2}** – 3 and 2x**^{3}** -5x + a, when divide by x-4**

ax^{3} + 3x^{2} – 3

r1 = a(4)^{3} + 3(4)^{2} – 3

= 64a + 48-3

= 64a + 45 ————–(i)

and

r2 = 2x^{3} -5x + a

= 2(4)^{3} -5(4) + a

= 128 – 20 +a

= 108 +a——————(ii)

but 2r1 = r2

2 x (64a + 45 ) = 108 +a

128a + 90 = 108 + a

128a – a = 108-90

127 a = 18

a = 18/127

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question-8. ****Using reminder theorem find the reminder obtain when x**^{3}** + ( kx+8) x + k, is divided by x+1 and x-2 hence find k if the sum of the two remainders is 1 (2019)**

**Answer:**

Given polynomial is p(x) = x^{3} +(kx + 8)x + k

g(x) = x + 1

∴ R_{1} = P(-1)

= (-1)^{3} + {k{-1} + 8} (-1) + k

= -1 + k – 8 + k

= 2k – 9

h(x) = x – 2

∴ R_{2} = P(2)

= (2)^{3} + (2k + 8)^{2} +k

= 8 + 4k + 16 + k

5k + 24

Now, R_{1} + R_{2} = 1

⇒ 2k – 9 + 5k + 24 = 1

⇒ 7k = 1 + 9 – 24

⇒ 7k = – 14

⇒ k = – 2

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question-9. ****By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x**^{2} + 5x – 3.

^{2}+ 5x – 3.

**Answer :**

Let x + 3 = 0 then x = – 3

Substituting the value of x in f(x)

f(-3) = (2 × (-3)^{2}) + (5 × -3) – 3

= (2 × 9) + (-15) – 3

= 18 – 15 – 3

= 18 – 18

= 0

Now, 2x – 1 = 0

Then, 2x = 1

x = ½

Given, f(x) = 2x^{2} + 5x – 3

Now, substitute the value of x in f(x),

f(½) = (2 × (½)^{2}) + (5 × ½) – 3

= (2 × (¼)) + 5/2 – 3

= ½ + 5/2 – 3

= (1 + 5)/2 – 3

= 6/2 – 3

= 3 -3

= 0

Reminder = 0 hence proved that, (x + 3) and (2x – 1) both are factors

### ML Aggarwal Ch-6 **Factorisation Exe-6 **Class 10 ICSE Maths Solutions

page-107

**Question -10. ****Without actual division, prove that X**^{4}+2x^{3}-2x^{2}+2x-3 is exactly divisible by x^{2} +2x – 3.

^{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).

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

^{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

x-2)3x²-x-10(3x+5

3x^{2}–6x

– +

———————

5x – 10

5x – 10

– +

———————

x

∴ 3x² – x – 10 = (x – 2)(3x + 5).

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question -12. ****Using the factor theorem, show that (x – 2) is a factor of x**^{3} + x^{2} – 4x – 4. Hence, factorise the polynomial completely. (2019)

^{3}+ x

^{2}– 4x – 4. Hence, factorise the polynomial completely. (2019)

**Answer:**

Given polynomial is p(x) = x^{3} + x^{2} – 4x – 4

x – 2 is its factor, if p(2) = 0

p(2) = (2)^{3} + (2)^{2} – 4(2) – 4 = 8 + 4 – 8 – 4 = 0

Thus, x – 2 is a factor of p(x).

Now, x^{3} + x^{2} – 4x + 4 = x^{2}(x +1) – 4(x + 1)

= (x + 1) (x^{2} – 4)

= (x + 1) (x + 2) (x – 2)

Hence, the required factors are (x + 1), (x + 2) and (x – 2).

**Question -13. ****Show that 2x + 7 is a factor of 2x**^{3} + 5x^{2} – 11x – 14. Hence factorise the given expression completely, using the factor theorem. (2006)

^{3}+ 5x

^{2}– 11x – 14. Hence factorise the given expression completely, using the factor theorem. (2006)

**Answer :**

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

x =

substituting the value of x in f(x),

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

= 2(-343/8) + 5(49/4) + (-77/2) – 14

= -343/4 + 245/4 – 77/2 – 14

= (-343 + 245 + 154 – 56)/4

= -399 + 399/4

= 0

Therefore, (2x + 7) is a factor of 2x^{3} + 5x^{2} – 11x – 14

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

So 2x^{3} + 5x^{2} – 11x – 14 can be written in product of divisor and quotient

= (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)

**Question-14 ****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

^{3}+ 2x

^{2}– 5x – 6

##### Factors of (∵ 6 = ± 1 : ± 2, ± 3, ±)

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)

**(ii) f(x) = x**^{3} – 13x – 12

^{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)]

= (x – 4)(x + 3)(x + 1)

**Question- 15. ****Use the Remainder Theorem to factorise the following expression **

(i) 2x^{3} + x^{2} – 13x + 6. (2010)

(ii) 3x^{2} + 2x^{2} – 19x + 6 (2012)

(iii) 2x^{3} +3x^{2} – 9x – 10. (2018)

(iv) x^{3} +10x^{2} – 37x + 26 (2014)

**Answer :**

**(i) Let f(x) = 2x**^{3} + x^{2} – 13x + 6

^{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

= (2 × 8) + 4 – 26 + 6

= 16 + 4 – 26 + 6

= 26 – 26

= 0

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

Dividing f(x) by x – 2,

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) Let f(x) = 3x**^{2} + 2x^{2} – 19x + 6

^{2}+ 2x

^{2}– 19x + 6

Using hit and trial method,

f(1) = 3 + 2 – 19 + 6 ≠ 0

f(−1) =–3 + 2 + 19 + 6 ≠ 0

f(2) = 24 + 8 – 38 + 6 = 0

Hence, (x – 2) is a factor of f(x)

**(iii) 2x**^{3} +3x^{2} – 9x – 10.

^{3}+3x

^{2}– 9x – 10.

Let p(x) = 2x^{3} + 3x^{2} – 9x – 10

Factors of constant term 10 are ± 1, ± 2, ± 5

Put x = 2, we have

p(2) =2(2)^{3} + 3(2)^{2} – 9(2) – 10

= 16 + 12 – 18 – 10

= 0

∴ (x – 2) is a factor of p(x)

Put x = – 1, we have

P(-1) =2(-1)^{3} + 3(-1)^{2} – 9 (-1) – 10

= – 2 + 3 + 9 – 10 = 0

∴ (x + 1) is a factor of p(x)

Thus, (x + 1) (x – 2) i.e.,x^{2} – x – 2 is a factor of p(x)

Hence, (x + 1), (x – 2) and (2x + 5) are the factors of given polynomial 2x^{3} + 3x^{2} – 9x – 10

**(iv) x**^{3} +10x^{2} – 37x + 26

^{3}+10x

^{2}– 37x + 26

**On dividing x ^{3} + 10x^{2} – 37x + 26 by x – 1, we get x^{2} + 11x – 26 as the quotient and remainder = 0.**

**Question -16. ****If (2 x + 1) is a factor of 6x**^{3} + 5x^{2} + ax – 2 find the value of a

^{3}+ 5x

^{2}+ ax – 2 find the value of a

**Answer :**

Let 2x + 1 = 0

Then, 2x = – 1

X = -½

Given, f(x) = 6x^{3} + 5x^{2} + ax – 2

substitute the value of x in f(x),

f (-½) = 6 (-½)^{3} + 5 (-½)^{2} + a (-½) – 2

= 6 (-1/8) + 5 (¼) – ½a – 2

= -3/4 + 5/4 – a/2 – 2

= (-3 + 4 – 2a – 8)/4

= (-6 – 2a)/4

, (2x + 1) is a factor of 6x^{3} + 5x^{2} + ax – 2

Then, remainder is 0.

So, (-6 – 2a)/4 = 0

-6 – 2a = 4 × 0

– 6 – 2a = 0

-2a = 6

a = -6/2

a = – 3

**Question-17. ****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

Given, f(x) = 3x^{3} – kx^{2} + 21x – 10

substitute the value of x in f(x),

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

= 3 (8/27) – k (4/9) + 14 – 10

= 8/9 – 4k/9 + 14 – 10

= 8/9 – 4k/9 + 4

= (8 – 4k + 36)/9

= (44 – 4k)/9

(3x – 2) is a factor of 3x^{3} – kx^{2} + 21x – 10

Then, remainder is 0

So, (44 – 4k)/9 = 0

44 – 4k = 0 × 9

44 = 4k

K = 44/4

K = 11

**Question -18. ****If (x – 2) is a factor of 2x**^{3} – x^{2} + px – 2, then

^{3}– x

^{2}+ px – 2, then

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

**(ii) with this value of p, factorize the above expression completely (2008)**

**Answer:**

**(i) Let x – 2 = 0, then x = 2**

Now f(x) = 2x^{3} – x^{2} + px – 2

substitute the value of x in f(x),

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

= (2 × 8) – 4 + 2p – 2

= 16 – 4 + 2p – 2

= 16 – 6 + 2p

= 10 + 2p

, (x – 2) is a factor of 2x^{3} – x^{2} + px – 2

Then, remainder is 0.

10 + 2p = 0

2p = – 10

P = -10/2

P = -5

So, (x – 2) is a factor of 2x^{3} – x^{2} + 5x – 2

##### (ii) 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 + 1) (x – 2) (2x + 1)

**Question -19. ****What number should be subtracted from 2x**^{3} – 5x^{2} + 5x so that the resulting polynomial has 2x – 3 as a factor?

^{3}– 5x

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

**Answer :**

Let P number to be subtracted from 2x^{3} – 5x^{2} + 5x

Then, f(x) = 2x^{3} – 5x^{2} + 5x – p

Given, 2x – 3 = 0

x = 3/2

f(3/2) = 0

So, f(3/2) = 2(3/2)^{3} – 5(3/2)^{2} + 5(3/2) – p = 0

2(27/8) – 5(9/4) + 15/2 – p = 0

27/4 – 45/4 + 15/2 – p = 0

27 – 45 + 30 – 4p = 0

57 – 45 – 4p = 0

12 – 4p = 0

P = 12/4

P = 3

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question -20**

(i) Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x^{3} + ax^{2} + bx – 12. (2001)

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

**Answer :**

**(i) Let x – 2 = 0**

Then, x = 2

Given, f(x) = x^{3} + ax^{2} + bx – 12

substitute the value of x in f(x),

f(2) = 2^{3} + a(2)^{2} + b(2) – 12

= 8 + 4a + 2b – 12

= 4a + 2b – 4

(x – 2) is a factor of x^{3} + ax^{2} + bx – 12.

So, 4a + 2b – 4 = 0

4a + 2b = 4

2a + b = 2 … (i)

Now, x + 3 = 0

Then, x = -3

Given, f(x) = x^{3} + ax^{2} + bx – 12

substitute the value of x in f(x),

f(-3) = (-3)^{3} + a(-3)^{2} + b(-3) – 12

= -27 + 9a – 3b – 12

= 9a – 3b – 39

(x – 3) is a factor of x^{3} + ax^{2} + bx – 12.

So, 9a – 3b – 39 = 0

9a – 3b = 39

3a – b = 13 … (ii)

, adding both equation (i) & (ii)

(2a + b) + (3a – b) = 2 + 13

2a + 3a + b – b = 15

5a = 15

a = 15/5

a = 3

Putting a = 3 inequation (i) to find ‘b’.

2a + b = 2

2(3) + b = 2

6 + b = 2

b = 2 – 6

b = -4

**(ii) **Let x + 2 = 0

Then, x = -2

Given, f(x) = x^{3} + ax + b

substitute the value of x in f(x),

f(-2) = (-2)^{3} + a(-2) + b

= -8 – 2a + b

(x + 2) is a factor of x^{3} + ax + b.

remainder is 0.

f(x) = 0

– 8 – 2a + b = 0

2a – b = – 8 … (i)

Let x + 3 = 0

Then, x = -3

Given, f(x) = x^{3} + ax + b

substitute the value of x in f(x),

f(-2) = (-3)^{3} + a(-3) + b

= -27 – 3a + b

(x + 3) is a factor of x^{3} + ax + b.

remainder is 0.

f(x) = 0

– 27 – 3a + b = 0

3a – b = – 27 … (i)

, subtracting

(2a – b) – (3a – b) = -8 – (-27)

2a – 3a – b + b = – 8 + 27

-a = 19

a = -19

Putting value of a = -19 inequation (i) to find ‘b’.

2a – b = – 8

2(-19) – b = -8

-38 – b = – 8

b = -38 +8

b = -30

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question-21. ****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 :**

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

Substituting the value of x in f(x),

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

substitute the value of x in f(x),

f(-2) = (-2)^{3} + a(-2) + b

= -8 – 2a + b

(x + 2) is a factor of x^{3} + ax + b.

remainder is 0.

f(x) = 0

– 8 – 2a + b = 0

2a – b = – 8 … (i)]

let x – 3 = 0

Then, x = 3

Given, f(x) = x^{3} + ax + b

, substitute the value of x in f(x),

f(3) = (3)^{3} + a(3) + b

= 27 + 3a + b

From the question, (x – 3) is a factor of x^{3} + ax + b.

Therefore, remainder is 0.

f(x) = 0

27 + 3a + b = 0

3a + b = – 27……… (ii)

adding both equation (i) and (ii)

(2a – b) + (3a + b) = – 8 – 27

2a – b + 3a + b = -35

5a = -35

a = -35/5

a = -7

putting a = -7 in equation (i) to find ‘b’.

2a – b = – 8

2(-7) – b = -8

-14 – b = -8

b = – 14 + 8

b = -6

value of a = -7 and b = -6.

Then, f(x) = x^{3} – 7x – 6

(x + 2) (x – 3)

= x(x – 3) + 2(x – 3)

= x^{2} – 3x + 2x – 6

= x^{2} – x – 6

Dividing f(x) by x^{2} – x – 6 we get,

x^{3} – 7x – 6 = (x + 1) (x + 2) (x – 3)

**Question -22. ****(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. (2005)

^{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. (2005)

**Answer :**

As x – 2 is a factor of

f(x) = x^{3} + ax^{2} + bx + 6 ……..(i)

Let x – 2 = 0

Then, x = 2

substitute the value of x in f(x),

f(2) = 2^{3}_{ }+ a(2)^{2} + 2b + 6

= 8 + 4a + 2b + 6

= 14 + 4a + 2b

= 7 + 2a + b

(x – 2) is a factor of the expression x^{3} + ax^{2} + bx + 6.

So, remainder is 0.

f(x) = 0

7 + 2a + b = 0

2a + b = -7 … (ii)

given that when expression is divided by (x – 3), it leaves the remainder 3.

remainder = 33 + 9a + 3b = 3

9a + 3b = 3 – 33

9a + 3b = -30

= 3a + b = – 10 … (iii)

subtracting equation (iii) from equation (ii)

(3a + b) – (2a + b) = – 10 – (-7)

3a – 2a + b – b = – 10 + 7

a = -3

Putting a = -3 equation (ii) to find ‘b’.

2a + b = – 7

2(-3) + b = – 7

-6 + b = – 7

b = – 7 + 6

b = – 1

**Question-23. ****If (x – 2) is a factor of the expression 2x**^{3} + ax^{2} + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. (2013)

^{3}+ ax

^{2}+ bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. (2013)

**Answer :**

f(x) = 2x^{3} + ax^{2} + bx – 14

∴ (x – 2) is factor of f(x)

f(2) = 0

f(2) = 2(2)^{3}_{ }+ a(2)^{2} + 2b – 14

= 16 + 4a + 2b – 14

= 2 + 4a + 2b

= 1 + 2a + b

R=0

2a + b = -1 … (i)

when expression is divided by (x – 3), it leaves the 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)

on Solving these (i) and (ii) simultaneous

2a + b = -1 … (i)

3a + b = 4 …(ii)

get value of a and b

a = 5

b= -2

**(ML Aggarwal Factorisation Exe-6 Class 10 ICSE)**

**Question -24. ****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.

^{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

Given, f(x) = ax^{3} + 3x^{2} + bx – 3

substitute the value of x in f(x),

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

= a(-27/8) + 3(9/4) – 3b/2 – 3

= -27a/8 + 27/4 – 3b/2 – 3

given that,** **ax^{3} + 3x^{2} + bx – 3 has a factor (2x + 3)**.**

remainder is 0.

-27a/8 + 27/4 – 3b/2 – 3 = 0

-27a + 54 – 12b – 24 = 0

-27a – 12b = -30

9a + 4b = 10 (i)

, let x + 2 = 0

Then, x = -2

Given, f(x) = ax^{3} + 3x^{2} + bx – 3

substitute the value of x in f(x),

f(2) = a(-2)^{3} + 3(-2)^{2} + b(-2) – 3

= -8a + 12 – 2b – 3

= -8a – 2b + 9

Leaves the remainder -3

So, -8a – 2b + 9 = -3

-8a – 2b = -3 – 9

-8a – 2b = -12

4a + b = 6 (ii)

By multiplying equation (ii) by 4,

16a + 4b = 24

Now, subtracting equation (ii) from equation (i) we get,

(16a + 4b) – (9a + 4b) = 24 – 10

16a – 9a + 4b – 4b = 14

7a = 14

a = 14/7

a = 2

put a = 2 equation (i) to find ‘b’.

9a + 4b = 10

9(2) + 4b = 10

18 + 4b = 10

4b = 10 – 18

4b = -8

b = -8/4

b = -2

Therefore, f(x) = ax^{3} + 3x^{2} + bx – 3

= 2x^{3} + 3x^{2} – 2x – 3

Given, 2x + 3 is a factor of f(x)

So, divide f(x) by 2x + 3

Therefore, 2x^{3} + 3x^{2} – 2x – 3 = (2x + 3) (x^{2} – 1)

= (2x + 3) (x + 1) (x – 1)

**Question- 25. ****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.

^{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:**

given that, f(x) = ax^{2} + bx + 2 and g(x) = bx^{2} + ax + 1 and x – 2 is a factor of f(x),

, x = 2

substitute the value of x in f(x),

f(2) = 0

a(2)^{2} + b(2) + 2 = 0

4a + 2b + 2 = 0

2a + b + 1 = 0 … (i)

when, g(x) divide by (x – 2), leaves remainder = – 15

g(x) = bx^{2} + ax + 1

So, g(2) = -15

b(2)^{2} + 2a + 1 = -15

4b + 2a + 1 + 15 = 0

4b + 2a + 16 = 0

2b + a + 8 = 0 … (ii)

Now, subtracting equation (ii) from equation (i) multiplied by 2,

(4a + 2b + 2) – (a + 2b + 8) = 0 – 0

4a – a + 2b – 2b + 2 – 8 = 0

3a – 6 = 0

3a = 6

a = 6/3

a = 2

let the equation (i) to find ‘b’.

2a + b + 1 = 0

2(2) + b = – 1

4 + b = – 1

b = – 1 – 4

b = – 5

Now, f(x) = ax^{2} + bx + 2 = 2x^{2} – 5x + 2

g(x) = bx^{2} + ax + 1 = -5x^{2} + 2x + 1

then, f(x) + g(x) + 4x^{2} + 7x

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

= x^{2} + 4x + 3

= x^{2} + 3x + x + 3

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

= (x + 1) (x + 3)

— : End of ML Aggarwal Factorisation Exe-6 Class 10 ICSE Maths Solutions Ch-6 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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