Question

The polynomials (ax^3+3x^2-3) and (2x^3-5x+a) when divided by (x-4) leave the same remainder,Find the value of a.

Answer 1

$a({4})^{3}+3({4})^{2}- 3= 2({4})^{3} \\ -5(4) + a \\ 64a + 48 - 3 = 128 - 20 + a$
$63a = 108 - 45\\ 63a = 63 \\ a = 1$

Answer 2

g(x)=x-4=0
g(x)=4


let ax³+3x²-3 be f(x)
and 2x³-5x+a be h(x)


f(x)=h(x)
as both have same remainder when decided by g(x)


a(4)³+3(4)²-3=2(4)³-5(4)+a
64a+48-3=128-20+a
64a+45=108+a
64a-a=108-45
63a=63
a=63÷63
=1
Hence the value of a is 1