How do you factor $8 z^{2} - 64$ $8z^2-64$ ?

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Answer 1 · 2017-03-29T01:30:07

See the entire solution process below:

$8$ $8$ is a common factor for both terms. Therefore, we can rewrite this expression as:

$8 z^{2} - 64 \to (8 \cdot z^{2}) - (8 \cdot 8) \to 8 (z^{2} - 8)$ $8z^2 - 64 -> (8 * z^2) - (8 * 8) -> 8(z^2 - 8)$

If necessary we can further factor the term: $z^{2} - 8$ $z^2 - 8$ .

We can use this rule to factor this term:

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a + b)(a - b)$

We can let $a^{2} = z^{2}$ $a^2 = z^2$ therefore $a = z$ $a = z$

We can let $b^{2} = 8$ $b^2 = 8$ therefore $b = \sqrt{8}$ $b = sqrt(8)$

Substituting gives:

$8 (z^{2} - 8) = 8 (z + \sqrt{8}) (z - \sqrt{8}$ $8(z^2 - 8) = 8(z + sqrt(8))(z - sqrt(8)$

How do you factor 8z2−648z^2-64?