RD Chapter 6 Factorisation of Polynomials Ex 6.4 Solutions
Question - 11 : - Find the value of a such that (x – 4) is a factor of5x3 – 7x2 – ax – 28.
Answer - 11 : -
Letf(x) = 5x3 – 7x2 – ax – 28
and Let x – 4 = 0, then x = 4
Now, f(4) = 5(4)3 – 7(4)2 – a x 4 – 28
= 5 x 64 – 7 x 16 – 4a – 28
= 320 – 112 – 4a – 28
= 320 – 140 – 4a
= 180 – 4a
∴ (x –4) is a factor of f(x)
∴Remainder = 0
⇒ 180 -4a = 0
⇒ 4a = 180
⇒ a = 1804 = 45
∴ a = 45
Question - 12 : - Find the value of a, if x + 2 is a factor of 4x4 +2x3 – 3x2 + 8x + 5a.
Answer - 12 : -
Letf(x) = 4x4 + 2x3 – 3x2 + 8x +5a
and Let x + 2 = 0, then x = -2
Now, f(-2) = 4(-2)4 + 2(-2)3 – 3(-2)2 +8 x ( 2) + 5a
= 4 x 16 + 2(-8) – 3(4) + 8 (-2) + 5a
= 64- 16- 12- 16 +5a
= 64 – 44 + 5a
= 20 + 5a
∴ (x + 2) is a factor of fix)
∴ Remainder = 0
⇒ 20 + 5a = 0 ⇒ 5a = -20
⇒ a =−205 = -4
∴ a = -4
Question - 13 : - Find the value of k if x – 3 is a factor of k2x3 –kx2 + 3kx – k.
Answer - 13 : -
Letf(x) = k2x3 – kx2 + 3kx – k
and Let x – 3 = 0, then x = 3
Now,f(3) = k2(3)3 – k(3)2 + 3k(3) –k
= 27k2 – 9k + 9k-k
= 27k2-k
∴ x – 3is a factor of fix)
∴Remainder = 0
∴ 27k2 –k = 0
⇒ k(27k– 1) = 0 Either k = 0
or 21k – 1 = 0
⇒ 21k =1
∴ k= 127
∴ k = 0,127
Question - 14 : - Find the values of a and b, if x2 – 4is a factor of ax4 + 2x3 – 3x2 +bx – 4.
Answer - 14 : -
f(x) =ax4 + 2x3 – 3x2 + bx – 4
Factors of x2 – 4 = (x)2 – (2)2
= (x + 2) (x – 2)
If x + 2 = 0, then x = -2
Now, f(-2) = a(-2)4 + 2(-2)3 – 3(-2)2 +b(-2) – 4
16a- 16 – 12-26-4
= 16a -2b-32
∵ x + 2is a factor of f(x)
∴Remainder = 0
⇒ 16a – 2b – 32 = 0
⇒ 8a – b– 16 = 0
⇒ 8a – b= 16 …(i)
Again x – 2 = 0, then x = 2
Now f(2) = a x (2)4 + 2(2)3 – 3(2)2 +b x 2-4
= 16a + 16- 12 + 26-4
= 16a + 2b
∵ x – 2 is a factor of f(x)
∴Remainder = 0
⇒ 16a + 2b = 0
⇒ 8a +b=0 …(ii)
Adding (i) and (ii),
⇒ 16a =16
⇒ a= 1616 =1
From (ii) 8 x 1 + b = 0
⇒ 8 + b= 0
⇒ b = – 8
∴ a = 1,b = -8
Question - 15 : - Find α and β, if x + 1 and x + 2 are factors of x3 +3x2 – 2αx +β.
Answer - 15 : -
Letf(x) = x3 + 3x2 – 2αx + β
and Let x + 1 = 0 then x = -1
Now,f(-1) = (1)3 + 3(-1)2 – 2α (-1) +β
= -1 + 3 + 2α + β
= 2 + 2α + β
∵ x + 1 is a factor of f(x)
∴ Remainder = 0
∴ 2 + 2α+ β = 0
⇒ 2α + β =-2 …(i)
Again, let x + 2 = 0, then x = -2
Now, f(-2) = (-2)3 + 3(-2)2 – 2α(-2) + β
= -8 + 12 + 4α+ β
= 4 + 4α+ β
∵ x + 2is a factor of(x)
∴Remainder = 0
∴ 4+ 4α+ β = 0
⇒ 4α + β = -4 …(ii)
Subtracting (i) from (ii),
2α = -2
⇒ α = −22 = -1
From (ii), 4(-1) + β = -4
-4 + β= -4
⇒ β =-4+ 4 = 0
∴ α = -1, β = 0
Question - 16 : - If x – 2 is a factor of each of the following two polynomials,find the values of a in each case:
(i) x3 – 2ax2 + ax – 1
(ii) x5 – 3x4 – ax3 + 3ax2 +2ax + 4
Answer - 16 : -
(i) Letf(x) = x3 – 2ax2 + ax – 1 and g(x) = x – 2
and let x – 2 = 0, then x = 2
∴ x – 2is its factor
∴Remainder = 0
f(2) = (2)3 – 2a x (2)2 + a x 2 – 1
= 8-8a+ 2a-1 = 7-6a
∴ 7 – 6a= 0
⇒ 6a = 7
⇒ a= 76
∴ a= 76
(ii) Let f(x) = x5 – 3x4 – ax3 +3 ax2 + 2ax + 4 and g(x) = x – 2
Let x – 2 = 0, then x=2
∴ f(2) =(2)5 – 3(2)4 – a(23) + 3a (2)2 +2a x 2 + 4
= 32 – 48 – 8a + 12a + 4a + 4
= -12 + 8a
∴Remainder = 0
∴ -12 +8a = 0
⇒ 8a= 12
⇒ a= 128 = 32
∴ Hencea = 32
Question - 17 : - In each of the following two polynomials, find thevalues of a, if x – a is a factor:
(i) x6 – ax5 + x4-ax3 +3x-a + 2
(ii) x5 – a2x3 + 2x + a + 1
Answer - 17 : - (i) Let f(x) = x6 –ax5+x4-ax3 + 3x-a + 2 and g(x) = x – a
∴ x – a is a factor
∴ x – a = 0
⇒ x = a
Now f(a) = a6-a x a5 + a4-a x a3 +3a – a + 2
= a6-a6 + a4-a4 + 2a + 2
= 2a + 2
∴ x + a is a factor of p(x)
∴ Remainder = 0
Question - 18 : - In each of the following, two polynomials, find thevalue of a, if x + a is a factor.
(i) x3 + ax2 – 2x + a + 4
(ii) x4 – a2r + 3x – a
Answer - 18 : -
Question - 19 : - Find the values of p and q so that x4 +px3 + 2x2 – 3x + q is divisible by (x2 –1).
Answer - 19 : -
Question - 20 : - Find the values of a and b so that (x + 1) and (x – 1)are factors of x4 + ax3 3x2 +2x + b.
Answer - 20 : -