How do you factor $g^{2} - 9$ $g^2-9$ ?

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Answer 1 · 2015-11-02T12:38:24

The factorised form for the expression is:
$(g + 3) (g - 3)$ $color(green)((g+3)(g-3)$

As per property:
$a^{2} - b^{2} = (a + b) (a - b)$ $color(green)(a^2-b^2=(a+b)(a-b)$

Applying the above to the expression provided:

$g^{2} - 9 = g^{2} - 3^{2}$ $g^2-9=g^2-3^2$

Here, $a = g, b = 3$ $color(green)(a)=g, color(green)(b)=3$

So,
$g^{2} - 9 = (g + 3) (g - 3)$ $g^2-9=color(green)((g+3)(g-3)$

How do you factor g2−9g^2-9 ?