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