# Thread: need help with factoring

1. ## need help with factoring

Can someone please explain to me how to do this problem. Thank You.

2x^2-3x-5=0

2. Originally Posted by hmonaghan09
Can someone please explain to me how to do this problem. Thank You.

2x^2-3x-5=0
the numbers here are easy enough to do trial and error with. you should have (2x + a)(x + b) = 0

now, you want a and b such that a*b = -5 and a + 2b = -3

or if trial and error isn't your thing. try the ac method.

multiply the a and c terms, that is the $2x^2$ and the $-5$, you get $-10x^2$

now you want two numbers that when you multiply them you get $-10x^2$ and when you add them you get the middle term ( $-3x$). note that $-5x$ and $+2x$ works.

split the middle term into these two numbers, you get

$2x^2 + 2x - 5x - 5 = 0$

factor the left side by grouping, you get

$2x(x + 1) - 5(x + 1) = 0$

$\Rightarrow (x + 1)(2x - 5) = 0$

now continue using the zero-factor property