How do you factor $x² - 4x - 45$ ?

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Answer 1 · 2015-07-06T21:34:58

Find two factors of $45$ whose difference is $4$ . The numbers $9$ and $5$ work.

Hence $x^2-4x-45 = (x-9)(x+5)$

$(x-a)(x+b) = x^2-(a-b)x-ab$

So given $x^2-4x-45$ , look for $a$ and $b$

such that $ab=45$ and $a-b = 4$ .

Answer 2 · 2015-07-06T21:51:45

Alternatively, complete the square to find:

$x^2-4x-45 = (x-2)^2-7^2 =$

$(x-2-7)(x-2+7) = (x-9)(x+5)$

$x^2-4x-45$

$=x^2-4x+4-4-45$

$=(x-2)^2-49$

$=(x-2)^2-7^2$

This is a difference of squares, so we can use the identity:

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

Let $a=x-2$ and $b=7$

Then

$(x-2)^2-7^2$

$=(x-2-7)(x-2+7) = (x-9)(x+5)$

How do you factor x² - 4x - 45x² - 4x - 45?