# Factorising

• April 5th 2006, 02:16 PM
raiden
Factorising

y = 2x^3 - x^2 - 8x + 4

also x + 2 is a factor
i used the remainder theorem and then long divided the equation
and i ended up with..

(x+2)(2x+1)(x-3)

help? lol
many thanks
• April 5th 2006, 07:56 PM
ThePerfectHacker
Quote:

Originally Posted by raiden

y = 2x^3 - x^2 - 8x + 4

also x + 2 is a factor
i used the remainder theorem and then long divided the equation
and i ended up with..

(x+2)(2x+1)(x-3)

help? lol
many thanks

Correction *Rayden not Raiden.

Anyways,
You have,
$2x^3-x^2-8x+4$
Group terms as,
$(2x^3-8x)-(x^2-4)$
Factor,
$2x(x^2-4)-(x^2-4)$
Factor again,
$(x^2-4)(2x-1)$
Factor again,
$(x+2)(x-2)(2x-1)$