Okay, really struggling on this one :/ I just need hep finishing the problem. x-1/(5x^2 - 8x +3) times 4(5x - 3)/x-11

- anonymous

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- schrodinger

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- anonymous

@zepdrix Let me know if I'm annoying you at anytime!!

- anonymous

@nincompoop please help, I'll fan & medal you (:

- zepdrix

you and your ways -_-

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## More answers

- zepdrix

What's going on?
Factoring stuff and simplify again?

- anonymous

Well, I'm just stuck here :/ I can't find a way to further factor it! I already factored a simplified some of it.

- zepdrix

Oh it's one of these types :O

- anonymous

Uh-oh, that doesnt sound good ;(

- zepdrix

|dw:1440828539698:dw|Have you learned this goofy method before?

- anonymous

Not at all!?

- zepdrix

Ehhh scratch that, let's do factor by GROUPING instead :)
That's a better approach anyway.

- anonymous

Ohh I know how to do that ;p But I thought you needed 4 terms?

- zepdrix

Hmm :d no.
You start with 3 terms, and yes, you'll break them into 4 terms in order to do your grouping.

- zepdrix

So what you're doing is, splitting the middle term in some clever way so you can have nice groups.

- zepdrix

\[\large\rm 5x^2\color{orangered}{-8x}+3\]

- anonymous

Okay, then I haven't done this before haha. Uhmm I really have no idea :/

- zepdrix

We'll multiply A and C, giving us 15.
Factors of 15 that add to -8?
Hmm

- zepdrix

15*1 = 15
15 + 1 = 16
Hmm those didn't work

- anonymous

5 and 3

- zepdrix

5 * 3 = 15
5 + 3 = 8
Hmm it didn't give us -8 though D:

- anonymous

;( I have no clue how to get -8 from the factors of 15..

- anonymous

Oh!!! Lol why do I always miss this?? -5 and -3! lol

- zepdrix

Ok good.
So that's telling us that we want to rewrite -8x as -5x and -3x.

- zepdrix

\[\large\rm 5x^2\color{orangered}{-8x}+3\]\[\large\rm 5x^2\color{orangered}{-5x-3x}+3\]

- anonymous

Ohh okay I remember this! And the groups are (5x^2 - 5x) and (-3x +3)

- zepdrix

And no, it doesn't matter which order you write them in.
If you had written -3x -5x in that order, we would end up with the same result.
So don't worry about that.

- zepdrix

Yes.
And please please please try to get in the habit of putting a `plus sign` between the groups. Very annoying mistake that a lot of students make lol

- zepdrix

(5x^2-5x) + (-3x+3)
k good :) now factor it out

- anonymous

Oh, sorry! So I got 5(x^2 - x) + 3(-x +1)

- zepdrix

MOAR :O
moar stuff

- anonymous

What do you mean?

- zepdrix

\[\large\rm 5(\color{orangered}{x^2 - x}) + 3(-x +1)\]Well, there is more in this orange stuff. The two terms in orange still share something in common.
Taking the 5 out was good, but there is moar.

- anonymous

Oh! Okay so it's 5x(x-1) + 3(-x+1)

- zepdrix

Mmmm k good.\[\large\rm 5x(\color{orangered}{x-1}) + 3(\color{orangered}{-x +1})\]But how do we get these orange parts to match up?
Hmm maybe we forgot to pull something out the second set of brackets?

- anonymous

5x(x-1) + -3(x+1) lol (:

- zepdrix

\[\large\rm \left[5x\color{orangered}{(x-1)}-3\color{orangered}{(x-1)}\right]\]OK good :O
We factored the two brackets completely.
I artificially placed some square brackets.
We now want to pull anything similar from both terms out of the square brackets.

- zepdrix

Hint Hint, I color-coded the thing we're pulling out :3

- anonymous

Lol so we just get 5x-3 (:

- zepdrix

\[\large\rm \left[5x\color{orangered}{(x-1)}-3\color{orangered}{(x-1)}\right]\quad=\quad\color{orangered}{(x-1)}\left[5x-3\right]\]Well we can't forget about the factor we pulled out, ya?
But yes, we end up with only 5x-3 in the square brackets.

- zepdrix

You don't have to use square brackets if you don't want :P
That was just help explain the process.
So our final result from factoring is:\[\large\rm 5x^2-8x+3\quad=\quad (x-1)(5x-3)\]

- zepdrix

Did the factoring from the square brackets step make sense? :o
I know that one can be a little tricky.

- anonymous

Ohhh my bad lol i just wasnt thinking, for some reason I thought the (x-1) just canceled out.. Lol so let me see if I can simplify now

- anonymous

Yayyyy! I solved it (: I got 4/x-11 thank you bunches

- zepdrix

yay \c:/

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