# cubic function factoring

• Sep 23rd 2007, 08:37 PM
ffezz
cubic function factoring
Hi. I have to factor a cubic function expression for homework and it just has me stomped.

x^3 - x^2 -5x -3

currently i have x+1 and x-3 as a factor (I did long division and got 0R for both of them but i cant figure out the third factor.

Appreciate any help
• Sep 23rd 2007, 08:49 PM
Jhevon
Quote:

Originally Posted by ffezz
Hi. I have to factor a cubic function expression for homework and it just has me stomped.

x^3 - x^2 -5x -3

currently i have x+1 and x-3 as a factor (I did long division and got 0R for both of them but i cant figure out the third factor.

Appreciate any help

what? i don't see how you got yourself in this predicament. after finding the one factor, finding the rest turns out to be finding the factors of a quadratic.

take the first factor and divide:

$x^3 - x^2 - 5x - 3 \div x + 1 = x^2 - 2x - 3$

$\Rightarrow x^3 - x^2 - 5x - 3 = (x + 1) \left( x^2 - 2x - 3 \right)$

$\Rightarrow x^3 - x^2 - 5x - 3 = (x + 1)(x - 3)(x + 1) = (x + 1)^2 (x - 3)$
• Sep 23rd 2007, 08:51 PM
ffezz
thanks =) i actually got that answer but for some reason it doesnt foil out to the original equation =x maybe doing math @ 2 in the morning after physics and chemistry is not a good idea. Thanks alot though
• Sep 23rd 2007, 08:56 PM
Jhevon
Quote:

Originally Posted by ffezz
maybe doing math @ 2 in the morning after physics and chemistry is not a good idea. Thanks alot though

nope, not a good idea. but sometimes you have to do what you have to do. like me, for instance, i have to pull an all-nighter tonight

you're welcome. good luck
• Sep 23rd 2007, 08:57 PM
ffezz
good luck =) il probably be back more in the future when i take calculus next semester lol :D