RD Chapter 6 Factorisation of Polynomials Ex MCQS Solutions
Question - 1 : - If x – 2 is a factor of x2 +3 ax – 2a, then a =
(a) 2
(b) -2
(c)1
(d) -1
Answer - 1 : -
∴ x – 2 is a factor of
f(x) = x2 + 3 ax – 2a
∴Remainder = 0
Let x – 2 = 0, then x = 2
Now f(2) = (2)2 + 3a x 2 – 2a
= 4 + 6a – 2a = 4 + 4a
∴Remainder = 0
∴ 4 + 4a = 0 ⇒ 4a =-4
⇒ a= −44 = -1
∴ a=-1 (d)
Question - 2 : - If x3 + 6x2 +4x + k is exactly divisible by x + 2, then k =
(a)-6
(b) -7
(c)-8
(d) -10
Answer - 2 : -
f(x) – x3 + 6x2 + 4x + k isdivisible by x + 2
∴Remainder = 0
Let x + 2 = 0, then x = -2
∴ f(-2) = (-2)3 + 6(-2)2 + 4(-2) + k
= -8 + 24-8 + k = 8 + k
∴ x + 2 is a factor
∴Remainder =0 .
⇒ 8 + k= 0 ⇒ k = -8
k =-8 (c)
Question - 3 : - If x – a is a factor of x3 –3x2 a + 2a2x + b, then the value of b is
(a)0
(b) 2
(c)1
(d) 3
Answer - 3 : -
∴ x – a is a factor of x3 – 3x2 a + 2a2x+ b
Let f(x) = x3 – 3x2 a + 2a2x+ b
and x – a = 0, then x = a
f(a) = a3 – 3a2.a + 2a2.a + b
= a3 – 3a3 + 2a3 + b = b
∵ x – a is a factor of f(x)
∴ b = 0 (a)
Question - 4 : - If x140 +2x151 + k is divisible by x + 1, then the value of k is
(a) 1
(b) -3
(c)2
(d) -2
Answer - 4 : -
∴ x + 1is a factor of f(x) = x140 + 2x151 + k
∴Remainder will be zero
Let x + 1 = 0, then x = -1
∴ f(-1) = (-1)140 + 2(-1)151 + k
= 1 + 2 x (-1) + k {∵ 140 is even and 151 is odd}
=1-2+k=k-1
∵Remainder = 0
∴ k –1=0 ⇒k=1 (a)
Question - 5 : - If x + 2 is a factor of x2 +mx + 14, then m =
(a)7
(b) 2
(c)9
(d) 14
Answer - 5 : -
x + 2is a factor of(x) = x2 + mx + 14
Let x + 2 = 0, then x = -2
f(-2) = (-2)2 + m{-2) + 14
= 4 – 2m + 14 = 18 – 2m
∴ x + 2 is a factor of f(x)
∴ Remainder = 0
⇒ 18 – 2m = 0
2m = 18 ⇒ m = 182 =9 (c)
Question - 6 : - If x – 3 is a factor of x2 –ax – 15, then a =
(a) -2
(b) 5
(c)-5
(d) 3
Answer - 6 : -
x – 3is a factor of(x) = x2 – ax – 15
Let x – 3 = 0, then x = 3
∴ f(3) =(3)2 – a(3) – 15
= 9 -3a- 15
= -6 -3a
∴ x – 3is a factor
∴ Remainder = 0
-6 – 3a = 0 ⇒ 3a = -6
∴ a = −63 = -2 (a)
Question - 7 : - If x51 + 51is divided by x + 1, the remainder is
(a)0
(b) 1
(c)49
(d) 50
Answer - 7 : -
Letf(x) = x51 + 51 is divisible by x + 1
Let x+1=0, then x = -1
∴ f(-1)= (-1)51 + 51 =-1+51 (∵ power 51 is an odd integer)
= 50 (d)
Question - 8 : - If x+ 1 is a factor of thepolynomial 2x2 + kx, then k =
(a)-2
(b) -3
(c) 4
(d) 2
Answer - 8 : -
∴ x + 1is a factor of the polynomial 2x2 + kx
Let x+1=0, then x = -1
Now f(x) = 2x2 + kx
∴Remainder =f(-1) = 0
= 2(-1)2 + k(-1)
= 2 x 1 + k x (-1) = 2 – k
∴ x + 1is a factor of f(x)
∴Remainder = 0
∴ 2 – k= 0 ⇒ k = 2 (d)
Question - 9 : - If x + a is a factor of x4 –a2x2 + 3x – 6a, then a =
(a)0
(b) -1
(c)1
(d) 2
Answer - 9 : -
x + a is a factor o f(x) = x4– a2x2 +3x – 6a
Let x + a = 0, then x = -a
Now, f(-a) – (-a)4 -a2(-a)2 + 3 (-a)– 6a
= a4-a4-3a-6a = -9a
∴ x + ais a factor of f(x)
∴Remainder= 0
∴ -9a =0 ⇒ a =0 (a)
Question - 10 : - The value of k for which x– 1 is a factor of 4x3 + 3x2 – 4x + k, is
(a) 3
(b) 1
(c)-2
(d) -3
Answer - 10 : -
x- 1 is a factor of f(x) = 4x3 + 3x2 –4x + k
Let x – 1 = 0, then x = 1
f(1) = 4(1 )3 + 3(1)2 – 4 x 1 + k
= 4+3-4+k=3+k
∴ x- 1is a factor of f(x)
∴Remainder = 0
∴ 3 + k= 0 ⇒ k =-3 (d)