How do you factor completely $a b^{2} - 3 b^{2} - 4 a + 12$ $ab^2 - 3b^2 - 4a + 12$ ?

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How do you factor completely $a b^{2} - 3 b^{2} - 4 a + 12$ $ab^2 - 3b^2 - 4a + 12$ ?

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Answer 1 · 2016-01-06T16:21:21

$= (b^{2} - 4) (a - 3)$ $=color(green)((b^2-4)(a-3)$

Explanation:

$a b^{2} - 3 b^{2} - 4 a + 12$ $ab^2 -3b^2-4a +12$

We can make groups of two to factorise this expression:
$(a b^{2} - 3 b^{2}) + (- 4 a + 12)$ $(ab^2 -3b^2) + (-4a +12)$

$b^{2}$ $b^2$ is common to both the terms in the first group, and $- 4$ $-4$ is common to both the terms in the second group.
We can write the expression as :
$b^{2} (a - 3) - 4 (a - 3)$ $b^2(a -3) -4 (a -3)$

Now, The term $(a - 3)$ $(a-3)$ is common to both the terms
$= (b^{2} - 4) (a - 3)$ $=color(green)((b^2-4)(a-3)$

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How do you factor completely $a b^{2} - 3 b^{2} - 4 a + 12$ $ab^2 - 3b^2 - 4a + 12$ ?

1 Answer

Explanation:

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How do you factor completely ab^2 - 3b^2 - 4a + 12ab2−3b2−4a+12ab^2 - 3b^2 - 4a + 12?

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How do you factor completely $a b^{2} - 3 b^{2} - 4 a + 12$ $ab^2 - 3b^2 - 4a + 12$ ?