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

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Answer 1 · 2015-05-28T01:14:10

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

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

The generic form of this equation is $a x^{2} + b x + c$ $ax^2+bx+c$ .

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

Multiply $a$ $a$ and $c$ $c$ .

$2 \cdot (- 3) = - 6$ $2*(-3)=-6$

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

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

Group the first and second pairs of terms.

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

Factor $x$ $x$ out of the first term.

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

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

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

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

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

Group the two sets of terms.

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

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

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

Factor out the common term $x + 1$ $x+1$ .

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

How do you factor 2x^2 - x - 32x2−x−32x^2 - x - 3?