Remainder Theorem - Factors of Polynomials

[Eklav Rai]

Given that 6x3 +5ax - 12a leaves a remainder of -4 when divided by x-a, find the possible values of a

[Open Intelligence]

Hello Eklav Rai,
We apologise for the delay in answering your question. Our maths tutors were taken up in the preparation of the forthcoming mock exams in August 2009.

(x-a) leaves a remainder of -4 means:
6a3 + 5a2 - 12a + 4 = 0
let f(x) = 6a3 + 5a2 - 12a + 4
by trial and error, when a = -2, the remainder is zero
Thus, (a+2) is a factor of f(x)

6a3 + 5a2 - 12a + 4 = (a+2)(pa2 + qa + c)
am using p,q and r instead of a, b and c simply to avoid confusion with the a that has been used in the question.

now compare coefficients for:
a3: 6 = p
constant: 4 = 2c
c = 2
a2: 5 = q + 2p
5=q+12
q=-7

Thus
(a+2)(6a2 - 7a + 2) = 0
(a+2)(3a-2)(2a-1)=0
a=-2,a=2/3 and a=1/2

[Eklav Rai]

Thanks for your help!!