Asked by Mike
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 (i) the values of constant p and q (ii) the zeros of f(x) (b) sketch the graph of f(x)
Answers
Answered by
oobleck
from the Remainder Theorem,
f(1) = 10
f(-1) = 12
so, now you have
2+p+q+18 = 10
-2+p-q+18 = 12
add to get
2p+36 = 22
p = -7
so q = -3
so f(x) = 2x^3-7x^2-3x+18 = (2x+3)(x-2)(x-3)
now finish it off
f(1) = 10
f(-1) = 12
so, now you have
2+p+q+18 = 10
-2+p-q+18 = 12
add to get
2p+36 = 22
p = -7
so q = -3
so f(x) = 2x^3-7x^2-3x+18 = (2x+3)(x-2)(x-3)
now finish it off
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