# How do you solve (x + 5)(3x+1)=0?

Aug 1, 2015

$\left\{\begin{matrix}{x}_{1} = - 5 \\ {x}_{2} = - \frac{1}{3}\end{matrix}\right.$

#### Explanation:

You can easily solve this equation by recognizing that the product of two terms is equal to zero if either one of those terms (or both, that's not the case here) is equal to zero.

In your case, the product of $\left(x + 5\right)$ and of $\left(3 x + 1\right)$ will be equal to zero when

$\left(x + 5\right) = 0$ or $\left(3 x + 1\right) = 0$

This means tha the two solutions to your equation will be

$x + 5 = 0 \implies {x}_{1} = \textcolor{g r e e n}{- 5}$

and

$3 x + 1 = 0 \implies {x}_{2} = \textcolor{g r e e n}{- \frac{1}{3}}$

If, for some reason, you'd want to complicate things a little, you could rewrite your equation in quadratic form

$\left(x + 5\right) \left(3 x + 1\right) = 0$

$3 {x}^{2} + 16 x + 5 = 0$

and use the quadratic formula to solve for the two solutions.

color(blue)(x_(1,2) = (_b +- sqrt(b^2 - 4ac))/(2a)

In your case, you have $a = 3$, $b = 16$, and $c = 5$, so

${x}_{1 , 2} = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{- 16 \pm \sqrt{196}}{6} = \frac{- 16 \pm 14}{8}$

Once again, you'd get

${x}_{1} = \frac{- 16 - 14}{6} = - 5$

and

${x}_{2} = \frac{- 16 + 14}{6} = - \frac{1}{3}$