• Feb 7th 2009, 03:03 AM
itsMoeyy

Please solve it for me and show me the correct and right method to solving these kind of questions.

(xcubed + 2xsquared - 5x - 6)

and Thanks alot :)
• Feb 7th 2009, 03:17 AM
Re :
Quote:

Originally Posted by itsMoeyy

Please solve it for me and show me the correct and right method to solving these kind of questions.

(xcubed + 2xsquared - 5x - 6)

and Thanks alot :)

\$\displaystyle P(x)=x^3+2x^2-5x-6\$

Well , are you familiar with the factor theorem ?
If not , note that for a polynomial P(x),(x-a) is a factor of P(x) if and only if P(a)=0 .

Notice that the constant term for the equation above is -6 .
Hence if (x-a) were to be a factor of P(x) , then a must be a factor of -6.
a=+/-1,+/-2,+/-3,+/-6

Substitute :

If x=1 , P(1)=1+2-5-6=-8 , hence (x-1) is NOT a factor .

If x=-1,P(-1)=-1+2+5-6=0 , hence (x+1) IS a factor .

Try doing this for all values of a stated above and you should get all the factors for this equation .
• Feb 7th 2009, 03:51 AM
itsMoeyy
Wow..

Thanks alot.. Your a genius! :D

But I have no idea how to do it..

Could you explain it a bit more?

and could you also solve the question? (Worried)

Im kind of dumb so yeahh.. =/
• Feb 7th 2009, 04:04 AM
Quote:

Originally Posted by itsMoeyy

Please solve it for me and show me the correct and right method to solving these kind of questions.

(xcubed + 2xsquared - 5x - 6)

and Thanks alot :)

\$\displaystyle
1 ------x^3+2x^2-5x-6)
\$
\$\displaystyle
2-----x^3+x^2+x^2+x-6x-6
\$
\$\displaystyle
3------x^2*(x+1)+x*(x+1)-6*(x+1)
\$
\$\displaystyle
4------(x^2+x-6)*(x+1)
\$
\$\displaystyle
5------(x^2+3x-2x-6)*(x+1)
\$
\$\displaystyle
6------(x*(x+3)-2*(x+3))*(x+1)
\$
\$\displaystyle
7------(x-2)*(x+3)*(x+1)
\$(Nod)
• Feb 7th 2009, 04:06 AM
itsMoeyy
HOLY MOLEY!

Niceee

How did ya do that? o.O
• Feb 7th 2009, 04:07 AM
Quote:

Originally Posted by itsMoeyy
Wow..

Thanks alot.. Your a genius! :D

But I have no idea how to do it..

Could you explain it a bit more?

and could you also solve the question? (Worried)

Im kind of dumb so yeahh.. =/

Erm , i have probably explained everything or perhaps i am missing something ..

but i think it is better if you post your working ( substitute all the a's like what i did ) or i do not know where else i can explain
• Feb 7th 2009, 05:04 AM