# How do you factor 3x^2-2x-8?

Feb 3, 2015

The solution is: $\left(3 x + 4\right) \left(x - 2\right)$

The polynomial is of the type $a {x}^{2} + b x + c$.

There are two ways to factor.

The first one:

we have to find two numbers whose sum is $- 2$ (that is $c$) and whose product is $- 24$ (that is $a \cdot c$).

$- 24 = - 1 \cdot 24$
$- 24 = - 24 \cdot 1$
$- 24 = - 2 \cdot 12$
$- 24 = - 12 \cdot 2$
$- 24 = - 3 \cdot 8$
$- 24 = - 8 \cdot 3$
$- 24 = - 4 \cdot 6$
$- 24 = - 6 \cdot 4$

The couple of numbers whose sum is $- 2$ is $- 6 , 4$.

Now we have to "split" the monomial $- 2 x$ in two parts: $- 2 x = - 6 x + 4 x$.

So our polynomial becomes:

$3 {x}^{2} - 6 x + 4 x - 8 = 3 x \left(x - 2\right) + 4 \left(x - 2\right) = \left(x - 2\right) \left(3 x + 4\right)$.

If it is known the way to solve an equation of 2° degree, here there is the second way to factor it:

We can imagine that this polynomial "becomes" an equation:

$3 {x}^{2} - 2 x - 8 = 0$, we can calculate the quantity

$\Delta = {b}^{2} - 4 a c = 4 + 96 = 100$,

and than we have to use this formula:

x_(1,2)=(-b+-sqrtDelta)/(2a)=(2+-10)/6rArrx_1=-4/3;x_2=2.

And now we have to use the formula:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.

So:

$3 {x}^{2} - 2 x - 8 = 3 \left(x + \frac{4}{3}\right) \left(x - 2\right) = \left(3 x + 4\right) \left(x - 2\right)$