How do you factor #a^ { 2} + 12a - 54= 10#?

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Answer 1 · 2017-03-27T06:32:54

Complete the square to find:

#(a-4)(a+16) = 0#

Note that:

#(a+6)^2 = a^2+12a+36#

So let us add #90# to both sides of the given equation to make the left hand side a perfect square:

#a^2+12a+36 = 100#

The resulting right hand side is a perfect square too and this can be written:

#(a+6)^2 = 10^2#

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

So with #A=(a+6)# and #B=10# we have:

#0 = (a+6)^2-10^2 = ((a+6)-10)((a+6)+10) = (a-4)(a+16)#

So we can write our original equation as:

#(a-4)(a+16) = 0#