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

Question

1140 views around the world

You can reuse this answer
Creative Commons License

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)#