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Dividing Polynomial difficult question

When

x^{3} + px^{2} + p^{2}x - 36

is divided by

x - 3

the remainder is 21. Find the possible values of P.

I got 2 and -5 but I did it in a REALLY long way. Is there a quickish way of doing this?

Reply 1

AwesomeNoise

did you do the f(x) has a factor (x-a) thing?

Reply 2

CHEM1STRY

Don't think so XD I am not quite sure what I did but it came with those answers... It's the only one I am unsure as to how to do quickly! It took me like a page of working

Reply 3

sango

Use the factor theorem.

So f(-3) = 0?

But f(-3) = 0 is when the original equation becomes -15 instead of -36 if I am right?

Reply 6

sango

CHEM1STRY

So f(-3) = 0?

No f(3) equals 0.
If x-3 is a factor then you need to make x-3=0 i.e. x should equal 3.

Reply 7

timotiis

The factor theorem: if (x - a) is a factor, then f(a) = 0.

So as (x - 3) is a factor, f(3) = 0

Reply 8

Swayum

No, f(3) = 23. f(3) = 0 would only work if the remainder was 0.

Please post in the right forum as well. This is for university course discussion, the academic help forum is linked to at the top and there's even a sticky thread in this forum!!!

Reply 9

mipegg

f(x)/f(x-a) will leave a remainder of f(a)...

Reply 10

CHEM1STRY

sango

No f(3) equals 0.
If x-3 is a factor then you need to make x-3=0 i.e. x should equal 3.

Oh yeah yeah I meant that sorry :biggrin:

Reply 11

Ze_Rebel_Mole

There's a remainder - so f(3) = 21

Reply 12

Swayum

mipegg

f(x)/f(x-a) will leave a remainder of f(a)...

timotiis

The factor theorem: if (x - a) is a factor, then f(a) = 0.

So as (x - 3) is a factor, f(3) = 0

sango

No f(3) equals 0.
If x-3 is a factor then you need to make x-3=0 i.e. x should equal 3.

sango

Use the factor theorem.

No, most of you need to read the question.

Reply 13

CHEM1STRY

Ze_Rebel_Mole

There's a remainder - so f(3) = 21

Ah Ok, thanks all it's easy now ! I never came accross the factor theorem and just realised I wasn't meant to do the question for HW... but anyway thanks alot for the information and I'll do it anyway now I know how