Question

Determine whether the following polynomials can be factored by grouping. If so, factor the polynomials showing all work. If not, explain why this method will not work.

2c2 + 4cd + 3c + 3d
4g2 + 12gh + 5g + 15h

Answer 1

@jim_thompson5910
@experimentX
@radar

Answer 2

is this it ?

Answer 3

@hba
@abb0t

Answer 4

um yah

Answer 5

finnaly abb0t is here :D

Answer 6

You can't combine like terms, but you CAN group them by their variables. so like c's together...

Answer 7

\(2c^2+3c +d(4c+3)\)

Answer 8

So it is unfactorable?

Answer 9

you can't factor out a common term from the whole function.

Answer 10

Okay so how could I express that in a way that my teacher would accept it xD

Answer 11

@JA1 got it
?

Answer 12

Ok so A is not factorable but B is?

Answer 13

yeah

Answer 14

Ok so how can I factor B ?

Answer 15

4g² + 12gh + 5g + 15h

Answer 16

\[2c ^{2} + 4cd + 3c + 3d\] Now I am going to group first two terms and last two terms and take the GCF out.

\[( 2c ^{2} + 4cd ) + (3c + 3d )\]

First 2 terms 2c is common. Last two terms 3 is common

\[2c ( c + 2d ) + 3 ( c + d)\]

But when I factor the GCF out the remaining factor should be the same in both groups. Which is not the same here. So no. 1 is not factorable.

Answer 17

Okay so what about no. 2?

Answer 18

itll be factored

Answer 19

yes but i need help with it

Answer 20

\[4g ^{2} + 12gh + 5g + 15h\] Try and group it like I did the first one.

Group the first two terms and last two terms and take the GCF out of each group. and show me your work

Answer 21

okeydokey

Answer 22

thanks for being my fan

Answer 23

what happened? I saw you typing something but nothing is posted

Answer 24

do it, show me the work, and I will be right back

Answer 25

\[(4g²+12gh) (5g+15h)\]

Answer 26

great

Answer 27

Now what is the GCF in the first group and GCF in the second group?

Answer 28

\[4g(g+3h) 5(q+3h)\]

Answer 29

like that?

Answer 30

\[(4g ^{2} + 12gh ) + ( 5g + 15h)\]

Answer 31

when you group it you keep the + sign in between the groups

Answer 32

ah ok

Answer 33

So
4g(g+3h)+ 5(q+3h)

Answer 34

now you did the GCF perfectly but you still have the + sign before the 5. So it will look like
\[4g ( g + 3h ) + 5 (g + 3h)\]

Answer 35

exactly

Answer 36

Now you have a common factor in both terms which is (g + 3h)

Answer 37

Now do distributive property in reverse

Answer 38

(4g + 5 ) ( g + 3h)

Answer 39

so thats the final factored equation?

Answer 40

yes

Answer 41

ah ok now i remember this, it was just in a different setting :D

Answer 42

Thanks to everyone