# How do you factor and solve 4x^2+8x= -3?

Jun 3, 2015

First, rewrite the equation as $4 {x}^{2} + 8 x + 3 = 0$.

Next, experiment with factorizations of the form $\left(x + a\right) \left(4 x + b\right)$ and $\left(2 x + c\right) \left(2 x + d\right)$ until you (hopefully) find something that works (note that you are forced to require that $a b = 3$ and $c d = 3$ so that you have a fairly limited number of choices for these numbers).

BTW, factorizations like this don't always work (the roots can be irrational or involve imaginary numbers), so you aren't guaranteed of success in general. However, "cookbook" textbook problems are usually chosen so that factorization is possible.

Eventually you should find that $4 {x}^{2} + 8 x + 3 = \left(2 x + 1\right) \left(2 x + 3\right)$.

That means the roots of the equation $4 {x}^{2} + 8 x + 3 = 0$ are the same as the roots of $\left(2 x + 1\right) \left(2 x + 3\right) = 0$. The only way this product can be zero is if one of the factors is zero, which implies $2 x + 1 = 0$ or $2 x + 3 = 0$, leading to $x = - \frac{1}{2}$ or $x = - \frac{3}{2}$. You can check that these both solve the original equation by substitution back into that equation:

$x = - \frac{1}{2}$: $L H S = 4 \cdot {\left(- \frac{1}{2}\right)}^{2} + 8 \left(- \frac{1}{2}\right) = 1 - 4 = - 3 = R H S$

$x = - \frac{3}{2}$: $L H S = 4 \cdot {\left(- \frac{3}{2}\right)}^{2} + 8 \left(- \frac{3}{2}\right) = 9 - 12 = - 3 = R H S$

("LHS" means left-hand side and "RHS" means right-hand side)