How do you factor the trinomial $x^{2} - 16$ $x^2 - 16$ ?

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How do you factor the trinomial $x^{2} - 16$ $x^2 - 16$ ?

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Answer 1 · 2015-12-03T23:58:06

$(x + 4) (x - 4)$ $(x+4)(x-4)$ .

Explanation:

We use $(a + b) (a - b)$ $(a+b)(a-b)$ for this type of operation. To do this, we have to find the way to put numbers squared. $x$ $x$ is already squared, but $16$ $16$ has to be put as $4^{2}$ $4^2$ . Now, $a = x$ $a = x$ and $b = 4$ $b = 4$ . So, we substitute:
$(x + 4) (x - 4)$ $(x+4)(x-4)$ .

Hope it Helps! :D .

Answer 2 · 2015-12-03T23:59:36

Apply the difference of squares formula to find that
$x^{2} - 16 = (x + 4) (x - 4)$ $x^2 - 16= (x+4)(x-4)$

Explanation:

The difference of squares formula states that
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a+b)(a-b)$
(try multiplying the right side out to verify this)

As $16 = 4^{2}$ $16 = 4^2$ we have

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

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How do you factor the trinomial $x^{2} - 16$ $x^2 - 16$ ?

2 Answers

Explanation:

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How do you factor the trinomial $x^{2} - 16$ $x^2 - 16$ ?