Question

Solve factoring
8x^2+20x=12

Answer 1

Hi
Welcome to opentudy :)
Any ideas on what we should do first?

Answer 2

I already divide by common number and turned it in to binomials
But I don't know what to do next

Answer 3

Ok show me what you have

Answer 4

I have gotten up to
(2x-1)(x+3)

Answer 5

Ok have you heard of the Zero Product property?

Answer 6

Yes

Answer 7

We would apply that here can you show me what 2 equations we would have when we use that property :)

Answer 8

Yes

Answer 9

Ok show me :-)

Answer 10

(2x-1)(x+3)=0

Answer 11

Correct :) so
\[\huge~\rm~\bf~2x−1=0 ~or~ x+3=0 \]
Solve both equations for x

Answer 12

X=1/2,x=-3

Answer 13

Good and those would be our answers

Answer 14

Omg thank u so much

Answer 15

yw :)

Answer 16

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Answer 17

X=1/2,x=-3

Answer 18

(2x-1)(x+3)=0

Answer 19

Yes

Answer 20

o.O

Answer 21

So,
8x^2+20x -12 = (2x-1)(x+3) ?

Answer 22

@wolf1728 we already solve for x.

Answer 23

changing to the quadratic form would've been easier and then factor out what's common in the 8, 20, and 12

Answer 24

I assumed they wanted to solve by factoring because of their post title.

Answer 25

Ya the question was to solve by factoring

Answer 26

Can someone help me on my new question I just posted

Answer 27

that equation wasn't in standard quadratic form though.
so subtracting 12 from both sides yields
\[8x^2+20x-12=0\]
and then 4 is what 8,20, and 12 have in common so factor 4 out
\[4(2x^2+5x-3)=0\]
\[4(2x-1)(x+3)=0\]
and then solve for x on each situation which is 1/2 and -3.

Answer 28

@UsukiDoll we did all of that ._.

Answer 29

standard quadratic form is ax^2+bx+c=0 ok then.. just showing that there is more than one way of solving this problem as long as none of the math rules are broken.

Answer 30

It's nice to know but the question had asked to solve by factoring

Answer 31

Ok now can someone Plzz help me with my new question I just posted

Answer 32

Can someone help me on my new question I just posted