How do you factor $x^{2} - 8 x + 16$ $x^2 - 8x + 16$ ?

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Answer 1 · 2018-07-06T14:09:01

$\Rightarrow {(x - 4)}^{2}$ $color(green)(=> (x-4)^2$

$x^{2} - 8 x + 16$ $x^2 - 8x + 16$

$\Rightarrow x^{2} - (2 \cdot 4 \cdot x) + 4^{2}$ $=>x^2 - (2 * 4 * x ) + 4^2$

It’s in the form $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2 - 2 ab + b^2 = (a- b)^2$

$:. => (x-4)^2$

Answer 2 · 2018-07-06T14:09:12

$(x-4)^2$

Given: $x^2-8x+16$ .

We find two numbers, say $a$ and $b$ , such that $a+b=-8$ and $ab=16$ .

Mentally, I see that $a=b=-4$ . Checking, $(-4)+(-4)=-8,-4*-4=16$ .

Now, we split it into:

$=x^2+ax+bx+16$

$=x^2-4x-4x+16$

$=x(x-4)-4(x-4)$

$=(x-4)(x-4)$

$=(x-4)^2$

How do you factor x^2 - 8x + 16x2−8x+16x^2 - 8x + 16?