# How do you solve 3x^2+11x-4=0 by factoring?

Mar 31, 2015

$3 {x}^{2} + 11 x - 4 = 0$

$a {x}^{2} + b x + c = 0$

Multiply $a c$ to get $- 12$

find factors the multiply to get $- 12$ and add to get the coefficient of the middle term $+ 11$

Because we want $- 12$, one factor is negative and the other is positive. Because we want the sum to be $+ 11$, the factor with greater absolute value is the positive factor:

List:
$- 1 \times 12$ sum $- 1 + 12 = 11$ STOP!, that's the one we want.

Now write the quadratic, replacing the middle term $11 x$ withe the two numbers we just found: $- x + 12 x$

$3 {x}^{2} - x + 12 x - 4 = 0$ Factor by grouping:

$\left(3 {x}^{2} - x\right) + \left(12 x - 4\right) = 0$

$x \left(3 x - 1\right) + 4 \left(3 x - 1\right) = 0$

$\left(x + 4\right) \left(3 x - 1\right) = 0$

$x + 4 = 0$ or $3 x - 1 = 0$

$x = - 4$ or $x = \frac{1}{3}$

The solutions are $- 4$ and $\frac{1}{3}$