# factorize

• December 26th 2008, 10:48 PM
sudha_14
factorize

1. Factorise:
a. $8x^2 - 2x - 15$
b. $2y^3 + y^2 - 2y - 1$
• December 26th 2008, 11:33 PM
nzmathman
By inspection,

$8x^2 - 2x - 15 = (2x - 3)(4x + 5)$

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.

$\text{Let} \,\,f(y) = 2y^3 + y^2 - 2y - 1$

Using the remainder theorem, we can see $f(1) = 2 + 1 - 2 - 1 = 0$
And therefore $y = 1$ is a root, therefore $(y - 1)$ is a factor of the polynomial.

Factorising: $2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1) = (y - 1)(2y + 1)(y + 1)$
• December 26th 2008, 11:48 PM
sudha_14
Quote:

Originally Posted by nzmathman
By inspection,

http://www.mathhelpforum.com/math-he...8fff355a-1.gif

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.

thanks A LOT! for the solution of the first equation....but for the second one...can you please show the long division..cuz even i understood HOW to do it....but i go wrong in the procedure....ALWAYS!!!! so i kinda needed help with that....
• December 27th 2008, 12:01 AM
nzmathman
Hmmm I know how to do the long division but cannot present it here nicely for you to see...I am not that advanced in the math coding system. I'm just able to factorise such expressions straight away...
• December 27th 2008, 12:03 AM
sudha_14
hmmm...sure .. no probz..ill try the long division again and if i still have a doubt..i'll tell u...
• December 27th 2008, 12:16 AM
Mathstud28
Quote:

Originally Posted by sudha_14

1. Factorise:
a. $8x^2 - 2x - 15$
b. $2y^3 + y^2 - 2y - 1$

There is no need for using polynomial division

\begin{aligned}2y^3+y^3-2y-1&=(2y^3+y^2)-(2y+1)\\
&=y^2(2y+1)-(2y+1)\\
&=(y^2-1)(2y+1)\\
&=(y-1)(y+1)(2y+1)\end{aligned}
• December 27th 2008, 12:16 AM
red_dog
b) Another approach, without the long division:
$2y^3+y^2-2y-1=y^2(2y+1)-(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)$
• December 27th 2008, 12:26 AM
sudha_14