How do you factor $x^{4} - 8 x^{2} + 16$ $x^4-8x^2+16$ ?

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How do you factor $x^{4} - 8 x^{2} + 16$ $x^4-8x^2+16$ ?

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Answer 1 · 2018-03-13T02:57:34

${(x^{2} - 4)}^{2}$ $(x^2-4)^2$

Explanation:

Make a dumby variable, in this case I will use $u$ $u$ .
Let $u = x^{2}$ $u=x^2$
You have $x^{4} - 8 x^{2} + 16$ $x^4-8x^2+16$ .
This is equivalent to ${(x^{2})}^{2} - 8 x^{2} + 16$ $(x^2)^2-8x^2+16$ .
You can now plug in $u$ $u$ wherever $x^{2}$ $x^2$ is.
You get $u^{2} - 8 u + 16$ $u^2-8u+16$ .
Now, you can factor this like a normal polynomial.
$(u - 4) (u - 4)$ $(u-4)(u-4)$
or
${(u - 4)}^{2}$ $(u-4)^2$ .
Finally, you plug in $u$ $u$ back into the expression.
The factored form will be ${(x^{2} - 4)}^{2}$ $(x^2-4)^2$ .

Answer 2 · 2018-03-13T02:58:18

${(x + 2)}^{2} {(x - 2)}^{2}$ $(x+2)^2(x-2)^2$

Explanation:

$= (x^{4} - 8 x^{2} + 16)$ $=(x^4-8x^2+16)$
$= (x^{2} - 4) (x^{2} - 4)$ $=(x^2-4)(x^2-4)$
$= (x + 2) (x - 2) (x + 2) (x - 2)$ $=(x+2)(x-2)(x+2)(x-2)$
$= {(x + 2)}^{2} {(x - 2)}^{2}$ $=(x+2)^2(x-2)^2$

Answer 3 · 2018-03-13T03:00:22

Notice how this is almost a simple quadratic trinomial. In fact, it can be thought of as a quadratic of $x^{2}$ $x^2$ .

Let $u = x^{2}$ $color(blue)(u = x^2)$

$x^{4} - 8 x^{2} + 16$ $x^4 - 8x^2 + 16$

$= {(x^{2})}^{2} - 8 (x^{2}) + 16$ $= (x^2)^2 - 8(x^2) + 16$

$= u^{2} - 8 u + 16$ $= color(blue)u^2 - 8color(blue)u + 16$

We can easily factor this quadratic.

$= {(u - 4)}^{2}$ $= (color(blue)u - 4)^2$

$= {(x^{2} - 4)}^{2}$ $= (color(blue)(x^2) - 4)^2$

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How do you factor $x^{4} - 8 x^{2} + 16$ $x^4-8x^2+16$ ?

3 Answers

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How do you factor $x^{4} - 8 x^{2} + 16$ $x^4-8x^2+16$ ?