How do you factor #-p^{2}+4p-4=0#?

1 Answer
Zachary
Mar 23, 2018

#-(p-2)^2=0#
This involves factoring out the negative from the trinomial.

Explanation:

You want the leading coefficient on the #p^2# to be positive. This means that you should factor out the negative, which will switch the signs.

#=># #-(p^2-4p+4)=0#

Now that the negative is factored out, you should factor the rest like normal. As #p^2# has no coefficient, the #p# in the factorization must have no coefficient. Seeing the #4#'s, 2 should pop into your head, because both #2+2# and #2*2# equal #4#. The constant is positive, which implies a #-2#, because #-2^2=4#.

These conclusions lead to:

#=># #(p-2)^2=0#

Now add the negative in front, for:

#=># #-(p-2)^2=0#

You can test the answer by using FOIL:

#=># #-((p-2)(p-2))=0#
#=># #-(p^2-4p+4)=0#

Distribute the negative:

#=># #-p^2+4p-4=0#

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