Remainder theorum

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• April 15th 2013, 02:25 AM
Andrew187
Remainder theorum
Determine the remainder when x³-6x²+x-5 is divided by:
a. X+2
b. X-3

Hi I was wondering if someone can help me with these revision questions.
• April 15th 2013, 03:06 AM
Plato
Re: Remainder theorum
Quote:

Originally Posted by Andrew187
Determine the remainder when x³-6x²+x-5 is divided by:
a. X+2
b. X-3.

If $f(x)=x^3-6x^2+x-5$ then if $f(a)=0$ then $(x-a)$ is a factor, so it divides $f(x)$.

Thus what is $f(-2)~\&~f(3)~?$
• April 15th 2013, 03:39 AM
Andrew187
Re: Remainder theorum
f(x) = x³ - 6x² + x - 5

(x+2)= f(-2) = (-2)³ - 6(-2)² + (-2) - 5 =

-8 - 24 - 2 - 5 = -39

Is this correct? I can't work out the answer for the first formula
• April 15th 2013, 04:01 AM
Prove It
Re: Remainder theorum
Quote:

Originally Posted by Plato
If $f(x)=x^2-6x^2+x-5$ then if $f(a)=0$ then $(x-a)$ is a factor, so it divides $f(x)$.

Thus what is $f(-2)~\&~f(3)~?$

That actually doesn't help the OP.

The remainder theorem states that for a polynomial function P(x) is divided by (x - a), the remainder is equal to P(a).
• April 15th 2013, 04:10 AM
Plato
Re: Remainder theorum
Quote:

Originally Posted by Andrew187
f(x) = x³ - 6x² + x - 5
(x+2)= f(-2) = (-2)³ - 6(-2)² + (-2) - 5 =
-8 - 24 - 2 - 5 = -39
Is this correct? I can't work out the answer for the first formula

Yes that is correct.
Thus $(x+2)$ does not divide $f(x)$. So $-39$ must be the remainder.

Surely you can find $(3)^3-6(3)^2+(3)-5=~?$
• April 15th 2013, 04:21 AM
Andrew187
Re: Remainder theorum
Quote:

Originally Posted by Plato
Yes that is correct.
Thus $(x+2)$ does not divide $f(x)$. So $-39$ must be the remainder.

Surely you can find $(3)^3-6(3)^2+(3)-5=~?$

its -29

But how come your formula is different to my first formula have I wrote it wrong?
• April 15th 2013, 04:25 AM
Plato
Re: Remainder theorum
Quote:

Originally Posted by Andrew187
But how come your formula is different to my first formula ?

How is it different?
• April 15th 2013, 04:45 AM
Andrew187
Re: Remainder theorum
Quote:

Originally Posted by Plato
Yes that is correct.
Thus $(x+2)$ does not divide $f(x)$. So $-39$ must be the remainder.

Surely you can find $(3)^3-6(3)^2+(3)-5=~?$

Is that sum for (x-3)?
• April 15th 2013, 04:55 AM
Plato
Re: Remainder theorum
Quote:

Originally Posted by Andrew187
Is that sum for (x-3)?

That is the remainder when $f(x)=x^3-6x^2+x-5$ is divided by $(x-3)$.
• April 15th 2013, 05:09 AM
Andrew187
Re: Remainder theorum
Quote:

Originally Posted by Plato
Yes that is correct.
Thus $(x+2)$ does not divide $f(x)$. So $-39$ must be the remainder.

Surely you can find $(3)^3-6(3)^2+(3)-5=~?$

27-24+3-5=1 is this correct?
or is it this 27-72+3-5= -43