How do you factor the expression $63cd^2 − 234cd + 135c$ ?

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Answer 1 · 2016-03-14T00:08:43

$63cd^2-234cd+135c=9c(7d-5)(d-3)$

All of the terms are divisible by $9c$ , so separate that out as a factor first:

$63cd^2-234cd+135c=9c(7d^2-26d+15)$

Factor the remaining quadratic expression using an AC method:

Look for a pair of factors of $AC=7*15=105$ with sum $B=26$

The pair $21, 5$ works.

Use that to split the middle term and factor by grouping:

$7d^2-26d+15$

$=7d^2-21d-5d+15$

$=(7d^2-21d)-(5d-15)$

$=7d(d-3)-5(d-3)$

$=(7d-5)(d-3)$

Putting it together:

$63cd^2-234cd+135c=9c(7d-5)(d-3)$

How do you factor the expression 63cd^2 − 234cd + 135c63cd^2 − 234cd + 135c?