Is it possible to factor #y=x^2 +8x +16 #? If so, what are the factors?

1 Answer
George C.
Jan 24, 2016

#x^2+8x+16 = (x+4)^2#

Explanation:

This is a perfect square trinomial.

First notice that both #x^2# and #16 = 4^2# are perfect squares.

Then multiply out #(x+4)^2# (using FOIL if you find it helpful) to see if the middle term matches:

#(x+4)^2#

#= (x+4)(x+4)#

#= stackrel "First" overbrace((x*x)) + stackrel "Outside" overbrace((x * 4)) + stackrel "Inside" overbrace((4 * x)) + stackrel "Last" overbrace((4*4))#

#=x^2+4x+4x+16#

#=x^2+8x+16#

In general note that #(a+b)^2 = a^2+2ab+b^2#, so if you recognise #a# and #b# then just check that the middle term is #2ab#.

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