Question

Factor completely:
3a^2-20ab+12b^2

Answer 1

\[3a ^{2}-20ab+12b ^{2}\]

Answer 2

multiply the first and last coefficients

3*12 = 36

Answer 3

find two numbers that multiply to 36
and
that add to -20

Answer 4

18 and 2

Answer 5

actually -18 and -2

-18 times -2 = +36
-18 plus -2 = -20

Answer 6

What is the next step? I'm confused on the different variables.

Answer 7

\[\Large 3a ^{2}\color{red}{-20ab}+12b ^{2}\]
\[\Large 3a ^{2}\color{red}{-18ab}\color{red}{-2ab}+12b ^{2}\]
now factor by grouping

Answer 8

(3a^2-18ab) (-2ab+12b^2)

Answer 9

don't forget the plus in between

(3a^2-18ab) + (-2ab+12b^2)

Answer 10

now factor each group

Answer 11

3a(a-6b) 2b(-1a+6b)

Answer 12

did i do it wrong? i feel like the 2nd group is incorrect.

Answer 13

(3a^2-18ab) + (-2ab+12b^2)

3a(a-6b) + (-2ab+12b^2)

3a(a-6b) -2b(a - 6b)

there's one more step

Answer 14

What would the next step be?

Answer 15

oh i know it :D

Answer 16

(3a-2b) (a-6b)

Answer 17

good

Answer 18

Thanks for the help @jim_thompson5910 :)

Answer 19

no problem