Question

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

Answer 1

`x=-1/3` and `x=-1`

Explanation:

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

`-6x-2=color(blue)((3x+1)(3x+1))`

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

• Multiply the first terms, `3x` and `3x` to get `color(blue)(9x^2)`
• Multiply the outside terms, `3x` and `1` to get `color(blue)(3x)`
• Multiply the inside terms, `1` and `3x` to get `color(blue)(3x)`
• Multiply the last terms, `1` and `1` to get `color(blue)(1)`

After foiling the right side binomial, we have:

`-6x-2=color(blue)(9x^2+3x+3x+1)`

`=>-6x-2=color(blue)(9x^2+6x+1)`

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

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

`9x^2+12x+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:

`9x^2+underbrace(9x+3x)_(12x)+3=0`

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

`color(red)(9x^2+9x)+color(purple)(3x+3)=0`

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

`underbrace(color(red)(9x)(x+1)+color(purple)(3)(x+1))_((9x+3)(x+1))=0`

`=>(9x+3)(x+1)=0`

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

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

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

We can solve these equations to get:

`x=-1/3` and `x=-1`

Hope this helps!