How do you factor $2x^2 - x - 3$ ?

Question

1702 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-28T01:14:10

$2x^2-x-3=(2x-3)(x+1)$

Problem: Factor $2x^2-x-3$ .

The generic form of this equation is $ax^2+bx+c$ .

$a=2$
$b=-1$
$c=-3$

Multiply $a$ and $c$ .

$2*(-3)=-6$

Find two factors of $-6$ that when added equal $-1$ . The numbers $-3$ and $2$ fit this requirement.

Rewrite the equation so that $-3x$ and $2x$ replace $-1x$ .

Group the first and second pairs of terms.

$(2x^2-3x)+(2x-3)$

Factor $x$ out of the first term.

$x(2x-3)+(2x-3)$ =

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

Factor out the common term $2x-3$ .

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

We can also rewrite the equation as $2x^2+2x-3x-3$ .

Group the two sets of terms.

$(2x^2+2x)-(3x+3)$

Factor $2x$ from the first term, and $3$ out of the second term.

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

Factor out the common term $x+1$ .

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

How do you factor 2x^2 - x - 32x^2 - x - 3?