How do you factor the trinomial $9(a + b)^2 - 60(a + b) + 100$ ?

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Answer 1 · 2016-07-23T11:46:30

${3(a+b)-10}^2$

$9(a+b)^2-60(a+b)+100$

= $9(a+b)^2-30(a+b) -30(a+b)+100$

= $3(a+b){3(a+b)-10} - 10{3(a+b)-10}$

= ${3(a+b)-10}{3(a+b)-10}$

= ${3(a+b)-10}^2$

Answer 2 · 2016-07-23T13:27:08

$(3(a+b)-10)(3(a+b)-10)$

= $(3(a+b)-10)^2$

This is simply a disguised quadratic which looks a whole lot worse than it is.

Let $(a + b)$ be $x$

The expression can now be written as:

$9x^2 -60x +100$

Find factors of 9 and 100 which add to give 60.

9 and 100 are both square numbers ... Let's work with that first.
Use factors of 9 and 100 and find the cross-product.

$" 3 10" rArr 3xx10 = 30$
$" 3 10" rArr 3xx10 = 30 " "30+30=60$

$(3x-10)(3x-10)$

But $x = (a+b)$

$(3(a+b)-10)(3(a+b)-10)$

How do you factor the trinomial 9(a + b)^2 - 60(a + b) + 100 9(a + b)^2 - 60(a + b) + 100?