# Remainder Theorem

• Nov 5th 2005, 02:45 AM
stormy_girl
Remainder Theorem
:confused:
Given that f(x) = x^2+px+q , determine the values of the constants p and q so that both
a) f(x) has a turning point when x=-3 and
b) the remainder when f(x) is divided by x+2 is 2.
• Nov 5th 2005, 06:59 PM
ticbol
Quote:

Given that f(x) = x^2+px+q , determine the values of the constants p and q so that both
a) f(x) has a turning point when x=-3 and
b) the remainder when f(x) is divided by x+2 is 2.
" a) f(x) has a turning point when x=-3 "
Turning point?
I assume that is a maximum or minimum point on the curve of f(x), so we equate f'(x) to zero:
f(x) = x^2 +px +q -----------(i)
f'(x) = 2x +p
Set f'(x) to zero, and x to -3,
0 = 2(-3) +p
p = 6 ---------***
So, for now,
f(x) = x^2 +6x +q --------(ii)

" b) the remainder when f(x) is divided by x+2 is 2. "
I cannot show the long division here. But (x^2 +6x +q) / (x+2) is
x+4, remainder (q-8).
So,
q-8 = 2
q = 2+8 = 10 ------***

Therefore, f(x) = x^2 +6x +10 -----------answer.
• Nov 5th 2005, 10:57 PM
rgep
Quote:

Originally Posted by ticbol
" b) the remainder when f(x) is divided by x+2 is 2. "
I cannot show the long division here.

No need to do the division. The remainder when polynomial f(x) is divided by (x-a) is just the value f(a). So the remainder when x^2 + 6x + q is divided by x+2 is (-2)^2 + 6(-2) + q = q-8.