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)?
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) = 16x3 – 8x2 + 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) = x3 – 7x2 + 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
= x2 – 4x + 3
∴ x3 – 7x2 +15x – 9
= (x – 3) (x2 – 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) = 2x3 + 3x2 – 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 3rd :
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 + x² – 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
∴ 2x3 – x2 – 5x – 2 = (x – 2) (2x2 + 3x + 1)
The expression can be the written as
(2x2 + 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.
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 x3 + 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 is10𝑥2)
(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) = 3x3 + 2x2 – 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 3x3 + 2x2 – 19x + 6 by x-2 we get,
3x2 + 8x – 3
∴ 3x3 + 2x2 – 19x + 6 = (x – 2)(3x2 + 8x – 3)
= (x – 2)(3x2 + 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) = 2x3 + x2 – 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 2x3 + x2 – 13x + 6 by x-2 we get
2x2 + 5x – 3
∴ 2x3 + x2 – 13x + 6 = (x – 2)(2x2 + 5x – 3)
= (x – 2)(2x2 + 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) = 2x3 – x2 – 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,
2x2 – 5x – 3
So, 2x3 – x2 – 13x – 6 = (x + 2)(2x2 – 5x – 3)
= (x + 2){2x2 – 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) = x3 + 2x2 – 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 + 2x2 –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.
x3 + 10x2 – 37x + 26
= (1)3 + 10(1)2 – 37(1) + 26
= 1 + 10 – 37 + 26
= 37 – 37
= 0
x – 1 is a factor of x3 + 10x2 – 37x + 26.
Now,
Divide x3 + 10x2 – 37x + 26 by x-1 we get,
x2 + 11x – 26
∴ x3 + 10x2 – 37x + 26
= (x – 1)(x2 + 11x – 26)
= (x – 1)(x2 + 13x – 2x – 26)
= (x – 1)[x(x + 13) – 2(x + 13)]
= (x – 1)(x + 13)(x – 2)
∴ x3 + 10x2 – 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) = 2x3 + 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
Please share with your friend
Thanks