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

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

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) = 2x3 + 3x2 – 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) = 2x3 + ax2 + 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.

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  R1
∴  f (-1) = R1
Also , f ( 2 ) = R2
∴  f (-1) = (-1)+ 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 :–

Return to:  ICSE Class 10 Maths RS Aggarwal Solutions

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