Question

i need help factoring these....

Answer 1

@austinL

Answer 2

Anyway, we have
\[\frac{(6x^2-32x+10)}{(3x^2-15x)} \div \frac{(3x^2+110x-4)}{(2x^2-32)}\]
Which is,
\[\frac{(6x^2-32x+10)}{(3x^2-15x)} \times \frac{(2x^2-32)}{(3x^2+11x-4)}\]
Lets begin with,
\[6x^2-32x+10\]

Answer 3

Go Austin!!! #swag

Answer 4

I don't know what to do

Answer 5

Ok, lets divide out the biggest factor that we can, what would that be?

Answer 6

2?

Answer 7

Right, so that gives us,
\[2(3x^2-16x+5)\]

Answer 8

okay.

Answer 9

So we now no that whatever factorization we get will be multiplied by two.
What would we put in the first spots of the pair?
2(___-?)(___-?)

Answer 10

2(x-?)(3x-?)... right?

Answer 11

okay

Answer 12

i don't know how to find the other two number still

Answer 13

So, we know we need a 16 and a 5, right?
So we know that we need a 5 in there. So, we have a 5 in the first blank, and a 1 in the second.

Answer 14

2(x-5)(3x-1)

Answer 15

lol okay

Answer 16

i don't know how to do the other 3 equations either

Answer 17

Ok... well... Do you want me to factor these out? Or do you want me to try and teach you? I am open to both ideas. It is whatever you want.

Answer 18

if you do them it might be faster i dont want you to spend too much time with it haha

Answer 19

Almost there.

Answer 20

When they are all completely factored we get this,
\[\frac{2(x-5)(3x-1)}{3(x-5)x} \times \frac{2(x-4)(x+4)}{(x+4)(3x-1)}\]

Answer 21

Can you take it from here?

Answer 22

yes! thanks! so you can have the x outside the parenthesis on the bottom of the first part?

Answer 23

You can indeed.

Answer 24

do you get 4(x-4)/3x as a final answer

Answer 25

Uno momento por favor.

Answer 26

I do get that as a final answer.

Answer 27

do you know how to factor x^2-5x