Question

The remainder when the polynomial f(x)=2x3+px2+qx+18 is divided by (x-1) is 10, when is divided by (x+1) the remainder is 12, find
(a) The values of p and q
(b) The zeros of f(x)

Answer 1

f(x)=2x^3+px^2+qx+18
f(1) = 2 + p + q + 18 = 10
p + q = -10

f(-1) = -2 + p - q + 18 = 12
p - q = -4

add the two p&q equations
2p = -14
p = -7, then in your head q = -3

for the roots
f(x)=2x^3 - 7x^2 - 3x + 18 , we already know x≠ ± 1
f(2) = 16 - 28 - 6 + 18 = 0 , YEAHH
so x= 2, and x-2 is a factor

by division:
2x^3 - 7x^2 - 3x + 18 = (x-2)(2x^2 - 3x - 9)
= (x-2)(x - 3)(2x + 3)

so the zeros are 2, 3, -3/2