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

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!