# Remainder theorum

• Apr 15th 2013, 01:25 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.
• Apr 15th 2013, 02:06 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 \$\displaystyle f(x)=x^3-6x^2+x-5\$ then if \$\displaystyle f(a)=0\$ then \$\displaystyle (x-a)\$ is a factor, so it divides \$\displaystyle f(x)\$.

Thus what is \$\displaystyle f(-2)~\&~f(3)~?\$
• Apr 15th 2013, 02:39 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
• Apr 15th 2013, 03:01 AM
Prove It
Re: Remainder theorum
Quote:

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

Thus what is \$\displaystyle 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).
• Apr 15th 2013, 03:10 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 \$\displaystyle (x+2)\$ does not divide \$\displaystyle f(x)\$. So \$\displaystyle -39\$ must be the remainder.

Surely you can find \$\displaystyle (3)^3-6(3)^2+(3)-5=~?\$
• Apr 15th 2013, 03:21 AM
Andrew187
Re: Remainder theorum
Quote:

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

Surely you can find \$\displaystyle (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?
• Apr 15th 2013, 03:25 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?
• Apr 15th 2013, 03:45 AM
Andrew187
Re: Remainder theorum
Quote:

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

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

Is that sum for (x-3)?
• Apr 15th 2013, 03:55 AM
Plato
Re: Remainder theorum
Quote:

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

That is the remainder when \$\displaystyle f(x)=x^3-6x^2+x-5\$ is divided by \$\displaystyle (x-3)\$.
• Apr 15th 2013, 04:09 AM
Andrew187
Re: Remainder theorum
Quote:

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

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

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