: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.

Printable View

- Nov 5th 2005, 03:45 AMstormy_girlRemainder 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, 07:59 PMticbolQuote:

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.

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, 11:57 PMrgepQuote:

Originally Posted by**ticbol**