How do you factor #(x+2)^2 -5(x+2)#?

2 Answers
Mar 17, 2018

Answer:

See a solution process below:

Explanation:

First, rewrite the term on the left as:

#(x + 2)(x + 2) - 5(x - 2)#

Next, factor out a common term of #(x - 2)# giving:

#((x + 2) - 5)(x - 2) =>#

#(x + 2 - 5)(x - 2) =>#

#(x - 3)(x - 2)#

Mar 17, 2018

Answer:

#(x-3)(x+2)#

Explanation:

First we expand the #(x+2)^2# which looks like #(x+2)(x+2)#

Next we just multiply it out which will give us
#x^2+4x+4#

Now that we know
#(x+2)^2 = x^2+4x+4#

We can do the other part which is #5(x+2)#
Which is #(5x+10)#
Now we know #5(x+2)=(5x+10)#

Now we can write the problem out expanded
#x^2+4x+4-(5x+10)#

NOTICE THE NEGATIVE SIGN IN FRONT OF THE PARENTHESES.
We have to distribute this negative to all terms in the parentheses.
Which gives us
#x^2+4x+4-5x-10#

Now we just combine like terms
#x^2-1x-6#

Now we need 2 numbers that multiply to #-6# and add up to #-1#
These numbers are #-3# and #2#

Notice #-3*2=-6#
And #-3+2=-1#

So we then get
#(x-3)(x+2)#