Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8B Goyal Brothers ICSE Maths Solutions Ch-8. Step by step solutions of exercise-8A questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

## Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8B Goyal Brothers ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-8 | Remainder and Factor Theorem |

Writer/ Book | RS Aggarwal |

Topics | Solution of Exe-8A Questions |

Academic Session | 2024-2025 |

### Factor Theorem

Factor theorem is commonly used for factoring a polynomial and finding the roots of the polynomial. It is a special case of a polynomial remainder theorem.

The factor theorem states that if f(x) is a polynomial of degree n greater than or equal to 1, and ‘a’ is any real number, then (x – a) is a factor of f(x) if f(a) = 0

#### How to Use Factor Theorem

Rather than finding the factors by using polynomial long division method, the best way to find the factors are factor theorem and synthetic division method

**Exercise- 8B Factor theorem**

**( Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-8 )**

**Que1: Using factor theorem, show that :**

**(i) (x-3) is a factor of (x³+x²-17x+15)**

**(ii) (x+1) is a factor of (x³+4x²+5x+2)**

**(iii) (3x-2) is a factor of (3x³+x²-20x+12)**

**(iv) (3-2x) is a factor of (2x³-9x²+x+12)**

**Sol: **(i) f(x) = (x³+x²-17x+15)

x-3 = 0

x = 3

f(3) = (3)³ + (3)² – 17(3) + 15

= 27+9-51+15

= 51-51 = 0.

Hence, (x-3) is a factor of f(x)

(ii) f(x) = (x³+4x²+5x+2)

x+1 = 0

x = -1

f(-1) = (-1)³ + 4(-1)² + 5(-1) + 2

= -1 + 4(1) -5 + 2

= 4-4 = 0

Hence, (x+1) is a factor of f(x).

(iii) f(x) = (3x³+x²-20x+12)

3x-2 = 0

x = 2/3

f(2/3) = 3(2/3)³ + (2/3)² – 20(2/3) + 12

= 3(8/27) + (4/9) – (40/3) + 12

= (8/9) + (4/9) – (40/3) + 12

= (8+4-120+108)/9

= (108-108)/9 = 0

Hence, (3x-2) is a factor of f(x).

(iv) f(x) = (2x³-9x²+x+12)

3-2x = 0

x = 3/2

f(3/2) = 2(3/2)³ – 9(3/2)² + (3/2) + 12

= 2(27/8) – 9(9/4) + (3/2) + 12

= (27/4) – (81/4) + (3/2) + 12

= (27-81+6+48)/12

= (81-81)/12 = 0.

Hence, (3-2x) is a factor of f(x).

**Que-2: Use factor theorem to show that (x+2) and (2x-3) are factors of (2x²+x-6).**

**Sol: **f(x) = (2x²+x-6)

g(x1) = x+2 = 0

x = -2

f(-2) = 2(-2)² + (-2) – 6

= 2(4) – 2 – 6

= 8-8 = 0

f(x) = (2x²+x-6)

g(x2) = 2x-3 = 0

x = 3/2

= 2(3/2)² + (3/2) – 6

= 2(9/4) + (3/2) – 6

= (9/2) + (3/2) – 6

= (9+3-12)/6

= (12-12)/6

= 0.

Hence both g(x) is a factor of f(x).

**Que-3: (i) Find the value of a so that (x+6) is a factor of the polynomial (x³+5x²-4x+a)**

(ii) For what value of a is the polynomial (2x³+ax²+11x+a+3) exactly divisible by (2x-1)?

(ii) For what value of a is the polynomial (2x³+ax²+11x+a+3) exactly divisible by (2x-1)?

**Sol: **(i) f(x) = (x³+5x²-4x+a)

x+6 = 0

x = -6

f(-6) = (-6)³ + 5(-6)² – 4(-6) + a

= -216 + 5(36) + 24 + a

= -216+180+24+a

= -12+a

a = 12.

(ii) f(x) = 2x³+ax²+11x+a+3

2x-1 = 0

x = 1/2

f(1/2) = 2(1/2)³ + a(1/2)² + 11(1/2) + a + 3

= 2(1/8) + a/4 + 11/2 + a + 3

= (1/4) + (a/4) + (11/2) + a + 3

-(a/4) – a = (1/4) + (11/2) + 3

(-a-4a)/4 = (1+22+12)/4

-5a = 35

a = 35/-5

a = -7.

**Que-4: What must be subtracted from 16x³-8x²+4x+7 so that the resulting expression has 2x+1 as a factor?**

**Sol: **Let the number to be subtracted from the given polynomial be k

Let f(x) = 16x^{3} – 8x^{2} + 4x + 7 – k

It is given that (2x + 1) is a factor of f(x).

∴ f(-12) = 0

⇒ 16(-1/2)³ – 8(-1/2)² + 4(-1/2) + 7 – k = 0

⇒ -16×(1/8) – 8×(1/4) – 4×(1/2) + 7 – k = 0

⇒ –2 – 2 – 2 + 7 – k = 0

⇒ – 6 + 7 – K = 0

⇒ 1 – k = 0

⇒ k = 1

Thus, 1 should be subtracted from the given polynomial.

**Que-5: Using factor theorem, show that (x-3) is a factor of (x³-7x²+15x-9). Hence factorise the given expression completely.**

**Sol: **Let p(x) = x^{3} – 7x^{2} + 15x – 9

For checking that (x – 3) is a factor of p(x), we find : p(3)

p(3) = (3)^{3} – 7(3)^{2} + 15(3) – 9

= 27 – 63 + 45 – 9

= 72 – 72

= 0.

Hence, (x – 3) is a factor of p(x).

By division of p(x) by x – 3, we get the quotient

= x^{2} – 4x + 3

∴ x^{3} – 7x^{2} +15x – 9

= (x – 3) (x^{2} – 4x + 3)

= (x – 3) (x – 3) (x – 1)

= (x – 3)^{2} (x – 1).

**Que-6: Using factor theorem, show that (x-4) is a factor of (2x³+x²-26x-40)and hence factorise (2x³+x²-26x-40).**

**Sol: **p(x) = 2x³+x²-26x-40

x-4 = 0

=> x = 4

Substituting 4 at place of x,

p(4) = 2(4)³+(4)²-26(4)-40 = 2(64)+16-104-40 = 128+16-104-40 = 0

The remainder is 0. Therefore, (x-4) is a factor of the polynomial.

2x³+x²-26x-40

= 2x³-8x²+9x²-36x+10x-40

= 2x²(x-4)+9x(x-4)+10(x-4)

= (x-4) {2x²+9x+10}

= (x-4) {2x²+4x+5x+10}

= (x-4) {2x(x+2)+5(x+2)}

= (x-4) (2x+5) (x+2)

**Que-7: Show that (x-3) is a factor of (2x³-3x²-11x+6) and hence factorise (2x³-3x²-11x+6).**

**Sol: **f(x) = 2x³-3x²-11x+6

x-3 = 0

x = 3

f(3) = 2(3)³ – 3(3)² – 11(3) + 6

= 2(54) – 3(9) – 33 + 6

= 54-27-33+6

= 60-60

= 0

Hence (x-3) is a factor of f(x).

2x³-3x²-11x +6.

2x³-6x²+ 3x²-9x-2x +6

= 2x²(x-3) + 3x (x-3) – 2 (x-3)

= (x-3) (2x²+3x-2)

= (x-3) (2x²+4x-x-2)

= (x-3) { 2x(x+2) – 1 (x+2)}

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

**Que-8: Show that (2x-3) is a factor of (2x³+3x²-5x-6) and hence factorise (2x³+3x²-5x-6).**

**Sol: **By remainder theorem we know that when a polynomial f(x) is divided by x – a, then the remainder is f(a).

Let f(x) = 2x^{3} + 3x^{2} – 5x – 6

f(2/3) = 2(3/2)^{3} + 3(3/2)^{2} – 5(3/2) – 6

= 2(27/8) + 3(9/4) – (15/2) – 6

= (54/8) + (27/4) – (15/2) – 6

= (54+54-60-48)/8

= (108-108)/8

= 0.

Thus, (x + 2) is a factor of the polynomial f(x).

According to the factor theorem , if for a polynomial , p ( x ) , p ( r ) is equal to 0 , ( x – r ) is a factor of P ( x ) .

If x = -1 ,

f ( x ) = 2x³ + 3x² – 5x – 6

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

=> f ( -1 ) = 0

=> ( x + 1 ) is a factor of 2x³ + 3x² – 5x – 6

Dividing 2x³ + 3x² – 5x – 6 by ( x – 1 )

=> 2x² + x – 6

Now ,

2x² + x – 6

=> 2x² + 4x – 3x – 6

=> 2x( x + 2 ) – 3 ( x + 2 )

=> ( 2x – 3 )( x + 2 )

Thus ,

p( x ) = 2x³ + 3x² – 5x – 6 can be Factorised as –

p ( x ) = ( 2x – 3 )( x + 2 )( x + 1 ) .

**Que-9: Show that (3x+2) is a factor of (6x³+13x²-4) and hence factorise (6x³+13x-4).**

**Sol: **f(x) = 6x³+13x²-4

If f ( x ) is a factor of p ( x ) , then p ( x ) = 0

Hence , lets put 3x + 2 = 0

3x = -2

x = – 2 / 3

Putting x = – 2 / 3 in given Expression , we get ;

= 6 * ( – 2 / 3 )³ + 13 ( – 2 / 3 )² – 4

= ( -16 / 9 ) + ( 52 / 9 ) – 4

= ( 52 – 16 ) / 9 – 4

= ( 36 / 9 ) – 4

= ( 4 – 4 ) = 0

Hence , when x = -2 / 3 , then p ( x ) = 0

Hence , by factor theorum , it justifies that ( 3x + 2 ) is a factor of ( 6x³ + 13x² – 4 ) .

Now ,

Divide p (x) by f (x) ,

( 6x³ + 13x² – 4 ) / ( 3x + 2 ) = 2x² + 3x – 2

=) 2x² + 3x – 2

=) 2x² + 4x – x – 2

=) 2x ( x + 2 ) **–**** **1 ( x + 2 )

=) ( 2x -1 )( x + 2 )

=) x = 1 / 2 , x = **–**2

Now ,

**Factor 1st :**

x = 1 / 2

x – ( 1 / 2 ) = 0

2x – 1 = 0 ( multiplying both side by 2 )

**Factor 2nd :**

x = -2

x + 2 = 0

**Factor ****3****r****d**** ****:
**x = – 2 / 3

x + ( 2 / 3 ) = 0.

3x + 2 = 0 ( multiplying both side by 3 )

Hence , we got that , ( 3x + 2 ) , ( x +2 ) & ( 2x -1 ) are the factors of 6x³ + 13x² – 4 .

**Que-10: Show that (2x+1) is a factor of (2x³+5x²+4x+1) and hence factorise (2x³+5x²+4x+1).**

**Sol: **f(x) = 2x³+5x²+4x+1

2x+1 = 0

x = -1/2

f(-1/2) = 2(-1/2)³ + 5(-1/2)² + 4(-1/2) + 1

= 2(-1/8) + 5(1/4) -2 + 1

= (-1/4) + (5/4) – 1

= (-1+5-4)/4

= (5-5)/4

= 0.

Now dividing 2x³+5x²+4x+1 by 2x+1, we get

(x²+2x+1)

Now, factor (x²+2x+1) we get,

(x+1)(x+1) = (x+1)²

f(x) = (2x+1)(x+1)²

**Que-11: Using the factor theorem, show that (x-2) is a factor of x³+x²-4x-4. Hence factorise the polynomial completely.**

**Sol: **f(x) = x³+x²-4x-4

Let x – 2 = 0, then x = 2

∴f(2) = (2)³+(2)²-4(2)-4

f (2) = 8 + 4 – 8 – 4

f (2) = 0

∴ x – 2 is a factor of f (x)

Now dividing x³ + x^{²} – 4x – 4 by x – 2, we get

= x³+x²-4x+4=(x-2)(x²+3x+2)

= (x – 2) (x + 2) (x + 1 ).

**Que-12: If (x-2) is a factor of 2x³-x²-px-2,**

(i) find the value of p

(ii) with the value of p, factorise the above expression completely.

**Sol: **(i) f(x) = 2x³-x²-px-2

x-2 = 0

x = 2

f(2) = 2(2)³ – (2)² – (2)p – 2

= 2(8) – (4) – 2p – 2

= 16-4-2p-2

2p = 10

p = 5.

(ii) After putting the value of p the equation become

f(x) = 2x³-x²-5x-2

Now divide 2x³-x²-5x-2 by (x-2) we get

2x²+3x+1

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

The expression can be the written as

(2x^{2} + 3x + 1) (x – 2) or (2x + 1) (x + 1) (x -2).

**Que-13: Find the value of a, if (x-a) is a factor of the polynomial 3x³+x²-ax-81.**

**Sol: **Let polynomial

P(x) = 3x³ + x² – ax – 81

x – a is a factor of P(x)

Then putting x – a = 0

i.e. x = a in P(x) we get

P(a) = 0

3a³ + a² – a × a – 81 = 0

3a³ = 81

a³ = 27

⇒ a = (27)^1/3 = 3

**Que-14: (i) Find the value of a and b, if (x-1) and (x+2) are both factors of (x³+ax²+bx-6).**

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

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

**Sol: **(i) f(x) = x³+ax²+bx-6

x-1 = 0

x = 1

1³+a(1)²+b(1)-6 = 0

1+a+b-6 = 0

a+b = 5 (i)

x+2 = 0

x = -2

(-2)³+a(-2)²+b(-2)-6 = 0

-8+ 4a -2b -6 = 0

4a-2b = 14

2(2a-b) = 14

2a-b = 7 (ii)

By comparing (I) and equation (ii)

a+b = 5

2a-b = 7

3a = 12

a = 4

Putting value of a in (i)

4 + b = 5

b = 5-4

b = 1

(ii) Given (x + 2) is a factor of x3+ax+b

⇒ (-2)³+a(-a)+b = 0 (x + 2 = 0 ⇒ x = -2)

⇒ -8-2a+b = 0

⇒ -2a+b = 8 …..(1)

Also, given that (x+ 3) is a factor of x^{3} + ax + b

⇒(-3)³+a(-3)+b = 0

⇒ -27-3a+b = 0

⇒ -3a+b = 27 …(2)

Subtracting (1) from (2) we have

-a = 19

⇒ a = -19

Substituting a = -19 in (1), we have

-2×(-19)+b = 8

⇒ 38+b = 8

⇒ b = -30`

Hence, a = -19 and b = -30

**Que-15: If (x³+ax²+bx+6) has (x-2) as a factor and leaves a remainder 3 when divided by (x-3), find the values of a and b.**

**Sol: **Let f(x) = x³ + ax² + bx + 6

By factor theorem we get f(2) = 0 [since x-2 = 0 then x= 2]

=> (2)³ + a(2)² + b(2) + 6 = 0

=>8 + 4a + 2b + 6 = 0

=> 4a + 2b= -14

=> 2(2a + b) = -14

=>2a + b = -7 ………….equation (1)

In the second case by remainder theorem we get, f(3) = 3 [since x-3= 0 then x= 3]

=>(3)³ + a(3)² + b(3) + 6 =3

=> 27 + 9a +3b + 6 = 3

=> 33 + 9a + 3b = 3

=> 9a + 3b = -30

=> 3(3a + b) = -30

=>3a + b = -10 ………….equation (2)

b= -7 -2a

Substituting the value of b in equation (2) we get,

3a -7 -2a= -10

=> a-7 = -10

=> a = -3

Substituting the value of a in equation (1) we get, 2(-3) +b = -7

=> -6 + b = -7

=>b = -1

Therefore the respective values of a and b are -3 and -1.

**Using Factor Theorem, Factrise each of the following :**

**Que-16: x³+7x²+7x-15**

**Sol: **x³ + 7x² + 7x – 15

x = 1

= 1³ + 7(1)² + 7(1) – 15

= 1 + 7 + 7 – 15

= 15 – 15

= 0

Hence x – 1 is one of the factor

Divide x³ + 7x² + 7x – 15 by x-1 we get,

(x² + 8x + 15)

x² + 8x + 15

= x² + 5x + 3x + 15

= x(x + 5) + 3(x + 5)

= (x + 5)(x + 3)

x³ + 7x² + 7x – 15 = (x-1)(x + 5)(x + 3)

**Que-17: 6x³-7x²-11x+12**

**Sol: **f(x) = 6x³-7x²-11x+12

Let x = 1

Putting the value in 6x³-7x²-11x+12

=6(1)³-7(1)²-11(1)+12

=6*1-7*1-11+12

=6-7-11+12

=18-18

=0

Divide 6x³-7x²-11x+12 by x-1 we get,

6x²-x-12

Factorise 6x²-x-12

6x²+8x-9x-12

2x(3x+4)-3(3x+4)

(2x-3)(3x+4)

6x³-7x²-11x+12 = (x-1)(2x-3)(3x+4).

**Que-18: 2x³+3x²-9x-10**

**Sol: **Let P(x) = 2×3+3×2-9x-10

P(2) = 16 + 12 – 18 – 10

P(2) = 0

So (x – 2) is a factor

Let us divide P(x) with (x-2), we get

(x-2)(2×2+7x+5)

This can be further factored to

(x-2)(2×2+5x+2x+5) ……… (Split 7x into two terms, whose sum is 7x and product is10𝑥^{2})

(x-2)(2×2+5x+2x+5)

(x-2))(x(2x+5)+1(2x+5))

(x – 2)(2x + 5)(x + 1)

**Que-19: 2x³+19x²+38x+21**

**Sol: **2x³ + 19x² + 38x + 21

=> 2x³ + ( 14 + 5 ) x² + ( 35 + 3 )x + 21

=> 2x³ + 14x² + 5x² + 35x + 3x + 21

=> 2x³ + 5x² + 15x + 14x² + 35x + 21

=> [ 2x³ + 5x² + 15x ] + [ 14x² + 35x + 21 ]

=> x [ 2x² + 5x + 3 ] + 7 [ 2x² + 5x + 3 ]

=> ( x + 7 )( 2x² + 5x + 3 )

=> ( x + 7 )( 2x² + 2x + 3x + 3 )

=> ( x + 7 )( 2x{ x + 1 } + 3 { x + 1 } )

=> ( x + 7 )( x + 1)( x + 3)

**Que-20: 3x³+2x²-19x+6**

**Sol: **Let P(x) = 3x^{3} + 2x^{2} – 19x + 6

By hit and trial method,

P(1) = 3(1)^{3} + 2(1)^{2} – 19(1) + 6

= 3 + 2 – 19 + 6

= –8 ≠ 0

P(–1) = 3(–1)^{3} + 2(–1)^{2} – 19(–1) + 6

= –3 + 2 + 19 + 6

= 24 ≠ 0

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

= 24 + 8 – 38 + 6

= 0

Thus, (x – 2) is a factor of P(x).

Now,

Divide 3x^{3} + 2x^{2} – 19x + 6 by x-2 we get,

3x^{2} + 8x – 3

∴ 3x^{3} + 2x^{2} – 19x + 6 = (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)

**Que-21: 2x³+x²-13x+6**

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

For x = 2

Factors of constant term 6 are ±1, ±2, ±3, ±6.

Putting x = 2, we have:

f(x) = f(2)

= 2(2)^{3} + (2)^{2} – 13(2) + 6

= 16 + 4 – 26 + 6

= 0

Hence, (x – 2) is a factor of f(x).

Now,

Divide 2x^{3} + x^{2} – 13x + 6 by x-2 we get

2x^{2} + 5x – 3

∴ 2x^{3} + x^{2} – 13x + 6 = (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)

**Que-22: 2x³-x²-13x-6**

**Sol: **Let f(x) = 2x^{3} – x^{2} – 13x – 6

By hit and trial method

Put x = – 2

f(–2) = 2(–2)^{3} – (–2)^{2} – 13(–2) – 6

= – 16 – 4 + 26 – 6

= 0

⇒ (x + 2) is a factor of f(x)

∴ Dividing f(x) by (x + 2)

We get,

2x^{2} – 5x – 3

So, 2x^{3} – x^{2} – 13x – 6 = (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)

**Que-23: If (x-2) is a factor of (x³+2x²-kx+10), find the value of k. Hence, determine whether (x+5) is also a factor of the given expression.**

**Sol: **p(x) = x^3 + 2x^2 – kx + 10

For (x – 2) to be the factor of p(x) = x^{3} + 2x^{2} – kx + 10

p(2) = 0

Thus (2)^{3} + 2(2)^{2} – k(2) + 10 = 0

⇒ 8 + 8 – 2k + 10 = 0

⇒ k = 13

Thus p(x) becomes x3 + 2x^{2} –13x + 10

Now, (x+5) would be the factor of p(x) iff p(–5) = 0

p(–5) = (–5)^{3} + 2(–5)^{2} – 13(–5) + 10 = –125 + 50 + 65 + 10 = 0

Thus, (x + 5) is also a factor of p(x).

**Que-24: Using the Remainder and Factor Theorems, factorise the polynomial. x³+10x²-37x+26.**

**Sol: **By remainder theorem,

For x = 1, the value of the given expression is the remainder.

x^{3} + 10x^{2} – 37x + 26

= (1)^{3} + 10(1)^{2} – 37(1) + 26

= 1 + 10 – 37 + 26

= 37 – 37

= 0

x – 1 is a factor of x^{3} + 10x^{2} – 37x + 26.

Now,

Divide x^{3} + 10x^{2} – 37x + 26 by x-1 we get,

x^{2} + 11x – 26

∴ x^{3} + 10x^{2} – 37x + 26

= (x – 1)(x^{2} + 11x – 26)

= (x – 1)(x^{2} + 13x – 2x – 26)

= (x – 1)[x(x + 13) – 2(x + 13)]

= (x – 1)(x + 13)(x – 2)

∴ x^{3} + 10x^{2} – 37 + 26 = (x – 1)(x + 13)(x – 2)

**Que-25: If (x-2) is a factor of the expression 2x³+ax²+bx-14 and when the expression is divided by (x-3), it leaves a remainder 52, find the values of a and b.**

**Sol: **Let p(x) = 2x^{3} + ax2 + bx – 14

Given, (x – 2) is a factor of p(x),

⇒ Remainder = p(2) = 0

⇒ 2(2)^{3} + a(2)^{2} + b(2) – 14 = 0

⇒ 16 + 4a + 2b – 14 = 0

⇒ 4a + 2b + 2 = 0

⇒ 2a + b + 1 = 0 …(1)

Given, when p(x) is divided by (x – 3), it leaves a remainder 52

∴ p(3) = 52

∴ 2(3)^{3} + a(3)^{2} + b(3) – 14 = 52

⇒ 54 + 9a + 3b – 14 – 52 = 0

⇒ 9a + 3b – 12 = 0

⇒ 3a + b – 4 = 0 …(2)

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

a – 5 = 0 ⇒ a = 5

From (1),

10 + b + 1 = 0 ⇒ b = –11

–: Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8B :–

Return to: ICSE Class 10 Maths RS Aggarwal Solutions

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