Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8A 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-8A 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 |

### Remainder Theorem

when a polynomial is p(a) is divided by another binomial (a – x), then the remainder of the end result that is obtained is p(x)

#### What is Remainder Theorem

The Remainder Theorem tells us that, in order to evaluate a polynomial *p*(*x*) at some number *x* = *a*, we can instead divide by the linear expression *x* − *a*. The remainder, *r*(*a*), gives the value of the polynomial at *x* = *a*

#### What is the remainder formula?

- Step 1: Divide the dividend by the divisor using integer division, ignoring any remainder.
- Step 2: Multiply the divisor by the quotient obtained in step 1.
- Step 3: Subtract the result obtained in step 2 from the dividend.
- Step 4: The resulting integer is the remainder.

**Exercise- 8A Remainder Theorem**

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

Using Remainder Theorem, find the remainder when :

**Que-1: f(x) = 3x²-5x+7 is divided by (x-2).**

**Sol: **x-2 = 0

x = 2

f(x) = 3x²-5x+7

f(2) = 3(2)²-5(2)+7

= 3(4)-10+7

= 12-3 = 9.

**Que-2: f(x) = 2x³-5x²+3x-10 is divided by (x-3).**

**Sol: **x-3 = 0

x = 3

f(x) = 2x³-5x²+3x-10

f(3) = 2(3)³ – 5(3)² + 3(3) – 10

= 2(27) – 5(9) + 9 – 10

= 54-45-1

= 8.

**Que-3: f(x) = 5x³-12x²+17x-6 is divided by (x-1).**

**Sol: **x-1 = 0

x = 1

f(x) = 5x³-12x²+17x-6

f(1) = 5(1)³ – 12(1)² + 17(1) – 6

= 5(1) – 12(1) + 17 – 6

= 5-12+11

= 4.

**Que-4: f(x) = x³-2x²-5x+6 is divided by (x+2).**

**Sol: **x+2 = 0

x = -2

f(x) = x³-2x²-5x+6

f(-2) = (-2)³ – 2(-2)² -5(-2) + 6

= -8 -2(4) + 10 + 6

= -8-8+16

= 0.

**Que-5: f(x) = 8x³-16x²+14x-5 is divided by (2x-1).**

**Sol: **2x-1 = 0

x = 1/2

f(x) = 8x³-16x²+14x-5

f(1/2) = 8(1/2)³ – 16(1/2)² + 14(1/2) – 5

= 8(1/8) – 16(1/4) + 7 – 5

= 1-4+2

= -1.

**Que-6: f(x) = 9x²-6x+2 is divided by (3x-2).**

**Sol: **3x-2 = 0

x = 2/3

f(x) = 9x²-6x+2

f(2/3) = 9(2/3)² – 6(2/3) + 2

= 9(4/9) – 4 + 2

= 4-2 = 2.

**Que-7: f(x) = 8x²-2x-15 is divided by (2x+3).**

**Sol: **2x+3 = 0

x = -3/2

f(x) = 8x²-2x-15

f(-3/2) = 8(-3/2)² – 2(-3/2) – 15

= 8(9/4) + 3 – 15

= 18-12 = 6.

**Que-8: On dividing (ax³+9x²+4x-10) by (x+3), we get 5 as remainder. Find the value of a.**

**Sol: **Let f(x) = ax³ + 9x² + 4x – 10

x + 3 = 0 ⇒ x = –3

On dividing f(x) by x + 3, it leaves a remainder 5.

∴ f(–3) = 5

a(–3)³ + 9(–3)² + 4(–3) – 10 = 5

–27a + 81 – 12 – 10 = 5

54 = 27a

a = 2

**Que-9: Using Remainder Theorem, find the value of k if on dividing 2x³+3x²-kx+5 by (x-2), leaves a remainder 7.**

**Sol: **Let f(x) = 2x^{3} + 3x^{2} – kx + 5

Using remainder theorem,

f(2) = 7

∴ 2(2)^{3} + 3(2)^{2} – k(2) + 5 = 7

⇒ 2(8) + 3(4) – k(2) + 5 = 7

⇒ 16 + 12 – 2k + 5 = 7

⇒ 2k = 16 + 12 + 5 – 7

⇒ 2k = 26

⇒ k = 13

**Que-10: If the polynomials (2x³+ax²+3x-5) and (x³+x²-2x+a) leaves the same remainder when divided by (x-2), find the value of a. Also, find the remainder in each case.**

**Sol: **If the two polynomial is same remainder means both are equal.

(2x³+ax²+3x-5) = (x³+x²-2x+a)

x-2 = 0

x = 2

2(2)³ + a(2)² + 3(2) – 5 = (2)³ + (2)²- 2(2) + a

= 2(8) + 4a + 6 – 5 = 8 + 4 – 4 + a

= 16+1+4a = 8+a

= 4a-a = 8-17

= 3a = -9

= a = -3.

f(x) = (2x³+ax²+3x-5)

f(2) = 2(2)³ + (-3)(2)² + 3(2) – 5

= 2(8) – 3(4) + 6 – 5

= 16-12+1

= 5.

f(x) = (x³+x²-2x+a)

f(2) = (2)³ + (2)² – 2(2) + (-3)

= 8+4-4-3

= 5. Hence Remainder in each case is 5.

**Que-11: The polynomials f(x) = (ax³+3x²-3) and g(x) = (2x³-5x+a) when divided by (x-4) leave the same remainder in each case. Find the value of a.**

**Sol: **If the two polynomial is same remainder means both are equal.

f(x) = g(x)

(ax³+3x²-3) = (2x³-5x+a)

x-4 = 0

x = 4

a(4)³ + 3(4)² – 3 = 2(4)³ – 5(4) + a

64a + 3(16) – 3 = 2(64) – 20 + a

64a+48-3 = 128-20+a

64a+45 = 108+a

64a-a = 108-45

63a = 63

a = 1.

**Que-12: Find a if the two polynomials ax³+3x²-9 and 2x³+4x+a leave the same remainder when divided by (x+3).**

**Sol: **If the two polynomial is same remainder means both are equal.

f(x) = g(x)

ax³+3x²-9 = 2x³+4x+a

x+3 = 0

x = -3

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

-27a + 3(9) – 9 = 2(-27) – 12 + a

-27a+27-9 = -54-12+a

-27a+18 = -66+a

18+66 = a+27a

84 = 28a

a = 3.

**Que-13: If (2x³+ax²+bx-2) when divided by (2x-3) and (x+3) leaves remainders 7 and -20 respectively, find the values of a and b.**

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

2x – 3 = 0

⇒ x = 3/2

On dividing f(x) by 2x – 3, it leaves a remainder 7.

= 2(3/2)³ + a(3/2)² + b(3/2) – 2 = 7

2(27/8) + a(9/4) + 3b/2 = 7+2

27/4 + 9a/4 + 3b/2 = 9

(27+9a+6b)/4 = 9

27 + 9a + 6b = 36

9a + 6b – 9 = 0

3a + 2b – 3 = 0 …(1)

x + 2 = 0 ⇒ x = –2

On dividing f(x) by x + 2, it leaves a remainder 0.

∴ 2(–2)^{3} + a(–2)^{2} + b(–2) – 2 = 0

–16 + 4a – 2b – 2 = 0

4a – 2b – 18 = 0 …(2)

Adding (1) and (2), we get,

7a – 21 = 0

a = 3

Substituting the value of a in (1), we get,

3(3) + 2b – 3 = 0

9 + 2b – 3 = 0

2b = –6

b = –3

**Que-14: Using the Remainder Theorem, find the remainder obtained when x³+(kx+8)x+k is divided by x+1 and x-2.**

Hence find k of the sum of two remainders is 1.

Hence find k of the sum of two remainders is 1.

**Sol: **Remainder theorem :

Dividend = Divisors × Quotient + Remainder

∴ Let f ( x ) = x³+(kx+8)x+k

=x³ +kx²+8x+k

Dividing f (x ) by x + 1 gives remainder as R_{1
}∴ f (-1) = R_{1
}Also , f ( 2 ) = R_{2
}∴ f (-1) = (-1)^{3 }+ k (-1)^{2} + 8 (-1) + k

= -1 + k – 8 + k

= 2k-9 = R1

f(2) = (2)³+k(2)²+8×2+k

= 8+4k+16+k

= 5k+24 = R2

Also, Sum of remainders = R1+ R1=1

∴ ( 2k – 9 ) + ( 5k +24 ) =1

7k + 15 = 1

7k = -14

k = – 2

**–: Remainder and Factor Theorem Class 10 RS Aggarwal Exe-8A Goyal Brothers ICSE Maths Solutions Ch-8 :–**

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