How do you factor completely #x^2 - 16x + 64#?

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Answer 1 · 2016-07-02T17:36:08

#(x -8)(x-8)#

The expression is a quadratic trinomial which is a product of two brackets.. #(x +- ?)(x +-?)#

In #x^2 color(blue)- 16x color(red)+64#
Find the factors of 64 which #color(red)(ADD)# up to 16.

The signs in the brackets will be #color(red)("THE SAME")#,

Both will be #color(blue)("MINUS")#.

#8 xx 8 = 64 and 8+8 = 16#

These are the factors we need.

#(x -8)(x-8)#

Answer 2 · 2016-07-02T17:39:51

#x^2-16x+64=(x-8)^2#

#x^2-16x+64#

= #x^2-8x-8x+64#

= #x(x-8)-8(x-8)#

= #(x-8)(x-8)#

= #(x-8)^2#