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

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

$x^2-16x+64$

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

= $x(x-8)-8(x-8)$

= $(x-8)(x-8)$

= $(x-8)^2$

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