RD Chapter 6 Factorisation of Polynomials Ex MCQS Solutions
Question - 11 : - If x+2 and x-1 are thefactors of x3+ 10x2 + mx + n, thenthe values of m and n are respectively
(a) 5and-3
(b) 17 and-8
(c) 7and-18
(d) 23 and -19
Answer - 11 : -
x+ 2and x – 1 are the factors of
f(x) = x3 + 10x2 + mx + n
Let x + 2 = 0, then x = -2
∴ f(-2)= (-2)3 + 10(-2)2 + m(-2) + n
= -8 + 40 – 2m + n = 32 – 2m + n
∴ x + 2 is a factor of f(x)
∴Remainder = 0
∴ 32 –2m + n = 0 ⇒ 2m – n = 32 …(i)
Again x – 1 is a factor of f(x)
Let x-1=0, then x= 1
∴ f(1) = (1)3 + 10(1)2 + m x 1 +n
= 1 + 10+ m + n =m + n+11
∴ x- 1 is a factor of f(x)
∴ m + n+11=0 ⇒ m+n=-11 …(ii)
Adding (i) and (ii),
3m = 32 – 11 = 21
⇒ m= 213 = 7
and n = -11 – m = m = 7, n = -18
∴ m= 7,n = -18 (c)
Question - 12 : - Let f(x) be a polynomial such that f( −12 )=0, then a factor of f(x) is
(a) 2x – 1
(b) 2x + b
(c) x- 1
(d) x + 1
Answer - 12 : -
Question - 13 : - When x3 –2x2 + ax – b is divided by x2 – 2x-3, theremainder is x – 6. The value of a and b are respectively.
(a) -2,-6
(b) 2 and -6
(c) -2 and6
(d) 2 and 6
Answer - 13 : -
Letf(x) = x3 – 2x2 + ax – b
and Dividing f(x) by x2 – 2x + 3
Remainder = x – 6
Let p(x) = x3 – 2x2 + ax – b – (x – 6) or x3 –2x2 + x(a – 1) – b + 6 is divisible by x2 – 2x+3 exactly
Now, x2-2x-3 = x2-3x + x- 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
∴ x – 3and x + 1 are the factors of p(x)
Let x – 3 = 0, then x = 3
∴ p(3) = (3)3 – 2(3)2 + (a-1)x3-b + 6
= 27-18 + 3a-3-b + 6
= 33-21+3 a-b
= 12 + 3 a-b
∴ x – 3is a factor
∴ 12 +3a-b = 0 ⇒ 3a-b =-12 …(i)
Again let x + 1 = 0, then x = -1
∴p(-1) =(-1)3 – 2(-l)2 + (a – 1) X (-1) – b + 6
= -1-2 + 1 – a – 6 + 6
= 4 – a- b
∴ x + 1 isa factor
4-a-b = 0 ⇒ a + b =4 …(ii)
Adding (i) and (ii),
4 a = -12 + 4 = -8 ⇒ a = −84 = -2
From (ii),
and -2 + 6 = 4⇒ 6 = 4+ 2 = 6
∴ a =-2, h =6 (b)
Question - 14 : - One factor of x4 +x2 – 20 is x2 + 5, the other is
(a) x2 –4
(b) x – 4
(c) x2-5
(d) x + 4
Answer - 14 : -
Question - 15 : - If (x – 1) is a factor ofpolynomial fix) but not of g(x), then it must be a factor of
(a) f(x)g(x)
(b) -f(x) + g(x)
(c) f(x) –g(x)
(d) {f(x) + g(x)}g(x)
Answer - 15 : -
∴ (x –1) is a factor of a polynomial f(x)
But not of a polynomial g(x)
∴ (x –1) will be the factor of the product of f(x) and g(x) (a)
Question - 16 : - (x + 1) is a factor of xn +1 only if
(a) n is an odd integer
(b) n is an eveninteger
(c) n is a negativeinteger
(d) n is a positiveinteger
Answer - 16 : -
∴ (x + 1) is a factor of xn+ 1
Let x + 1 = 0, then x = -1
∴ f(x) =xn + 1
and f(-1) = (-1)n + 1
But (-1)n is positive if n is an even integer and negative if nis an odd integer and (-1)n +1=0 {∵ x + 1 is a factor of(x)}
(-1)n must be negative
∴ n isan oddinteger (a)
Question - 17 : - If x2 + x +1 is a factor of the polynomial 3x3 + 8x2 + 8x+ 3 + 5k, then the value of k is
(a) 0
(b) 25
(c) 52
(d) -1
Answer - 17 : - x2 + x + 1 is a factor of
f(x) = 3x3 + 8x2 + 8x + 3 + 5k
Now dividing by x2 + x + 1, we get
Question - 18 : - If (3x – 1)7 =a7x7 + a6x6 + a5x5 +… + a1x + a0, then a7 + a6 +a5 + … + a1 + a0 =
(a)0
(b) 1
(c) 128
(d) 64
Answer - 18 : -
f(x) = [3(1) – 1]7 = a7x7 +a6x6 + a5x5 + … + a1x+ a0
Let x = 1, then
f(1) = (3x- 1)7 = a7(1)7 + a6(1)6 +a5(1)5 + … + a1 x 1 + a0
⇒ (3 – 1)7 = a7 x 1 + a6 x 1+ a5 x1 + … + a1x 1 +a0
⇒ (2)7 = a7 + a6 + a5 + … + a1 +a0
∴ a7 +a6 + a5 + … + a1 + a0=128 (c)
Question - 19 : - If both x – 2 and x – 12 are factors of px2 + 5x + r,then
(a) p =r
(b) p + r = 0
(c) 2p + r =0
(d) p + 2r = 0
Answer - 19 : -
Question - 20 : - If x2 – 1is a factor of ax4 + bx3 + cx2 +dx + e, then
(a) a + c + e- b + d
(b)a + b + e = c + d
(c)a + b + c = d+ e
(d)b + c + d= a + e
Answer - 20 : -
X2 – 1 is a factor of ax4 + bx3 +cx2 + dx + e
⇒ (x + 1), (x – 1) are the factors of ax4 + bx3 +cx2 + dx + e
Let f(x) = ax4 + bx3 + cx2 + dx+ e
and x + 1 = 0 then x = -1
∴ f(-1) = a(-1)4 + b(-1)3 + c(-1)2 +d(-1) + e
= a- b + c- d+ e
∴ x + 1is a factor of f(x)
∴ a-b + c- d+e = 0
⇒ a + c+ e = b + d (a)