# How do you solve -6x - 2 = (3x + 1)^2?

Mar 18, 2018

$x = - \frac{1}{3}$ and $x = - 1$

#### Explanation:

Let's start with the right-hand side. We can rewrite the equation as:

$- 6 x - 2 = \textcolor{b l u e}{\left(3 x + 1\right) \left(3 x + 1\right)}$

What I have in blue, we can multiply using the highly useful mnemonic FOIL:

• Multiply the first terms, $3 x$ and $3 x$ to get $\textcolor{b l u e}{9 {x}^{2}}$
• Multiply the outside terms, $3 x$ and $1$ to get $\textcolor{b l u e}{3 x}$
• Multiply the inside terms, $1$ and $3 x$ to get $\textcolor{b l u e}{3 x}$
• Multiply the last terms, $1$ and $1$ to get $\textcolor{b l u e}{1}$

After foiling the right side binomial, we have:

$- 6 x - 2 = \textcolor{b l u e}{9 {x}^{2} + 3 x + 3 x + 1}$

$\implies - 6 x - 2 = \textcolor{b l u e}{9 {x}^{2} + 6 x + 1}$

This is a quadratic equation, so we want to set it equal to zero to find its zeros.

Let's add $6 x$ and $2$ to both sides of this equation to get

$9 {x}^{2} + 12 x + 3 = 0$

Now, we can factor by grouping. This is splitting up the $b$ term so that we can be able to factor. This can be rewritten as:

$9 {x}^{2} + {\underbrace{9 x + 3 x}}_{12 x} + 3 = 0$

Notice, $9 x + 3 x = 12 x$, so I have not changed the value of this equation.

$\textcolor{red}{9 {x}^{2} + 9 x} + \textcolor{p u r p \le}{3 x + 3} = 0$

We can factor a $9 x$ out of the red term, and a $3$ out of the purple term. We get:

${\underbrace{\textcolor{red}{9 x} \left(x + 1\right) + \textcolor{p u r p \le}{3} \left(x + 1\right)}}_{\left(9 x + 3\right) \left(x + 1\right)} = 0$

$\implies \left(9 x + 3\right) \left(x + 1\right) = 0$

NOTE: I was able to rewrite this as the product of two binomials because both terms had an $\left(x + 1\right)$ in common.

Now we can set our two binomials equal to zero. We get:

$9 x + 3 = 0$ and $x + 1 = 0$

We can solve these equations to get:

$x = - \frac{1}{3}$ and $x = - 1$

Hope this helps!