Can somebody help me figure out by factoring? while listing the steps so I can understand how you did it? Thanks!:)
3(5-x)^4+2(5-x)^3-(5-x)^2

- anonymous

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- zepdrix

Hey :)
\[\large\rm 3(5-x)^4+2(5-x)^3-(5-x)^2\]

- zepdrix

\[\large\rm 3(5-x)(5-x)(5-x)^2+2(5-x)(5-x)^2-(5-x)^2\]The first one is a 4th power,
so I can pull a couple of them out
(this isn't the factoring,
I'm just trying to get a feel for what I'll be factoring out of each term.
I'm just applying this idea right now: \(\large\rm x^4=x\cdot x\cdot x^2\))

- zepdrix

From here we can clearly see what they have in common:\[\large\rm 3(5-x)(5-x)\color{#DD4747}{(5-x)^2}+2(5-x)\color{#DD4747}{(5-x)^2}-\color{#DD4747}{(5-x)^2}\]

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## More answers

- zepdrix

Oh let's add the brackets on the outside of everything,
maybe that will make things a little more clear.\[\large\rm \left[3(5-x)(5-x)\color{#DD4747}{(5-x)^2}+2(5-x)\color{#DD4747}{(5-x)^2}-\color{#DD4747}{(5-x)^2}\right]\]So we're doing is,
we're pulling this common factor out of the square brackets.\[\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-?\right]\]Gotta be careful here.
What am I going to get in my last term? :)
When you fully factor something out of itself.

- anonymous

why did you keep (5-x)^2 at the end? why not just write it out another time?

- anonymous

is that a GCF?

- zepdrix

Yes.
It appears that (5-x)^2 was the GCF!
So we didn't really want to break it down into (5-x)(5-x),
you could though :)

- anonymous

okaayyyy i got it, now would the -? at the end be -1?

- zepdrix

Good!
When you divided (5-x)^2 out of (5-x)^2, you're left with a 1.
You're not left with a 0,
I just wanted to make sure that was clear.\[\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-1\right]\]

- anonymous

okay so now what? can it be simplified further?

- zepdrix

\[\large\rm =(5-x)^2\left[\color{royalblue}{3(5-x)(5-x)+2(5-x)-1}\right]\]
`Maybe`.
We'll have to do the work of expanding out all of this blue nonsense before we can determine whether it can be factored further.

- zepdrix

Do you understand how to multiply (5-x)(5-x)? :)
Think you can get the blue stuff into the form ax^2+bx+c?

- anonymous

I understand that they are binomials and can be simplified by using FOIL correct? or do we not use this equation for this?

- zepdrix

Good :)
We've done the GCF method already, there isn't any more scraps to pull out of there.
Now we have to rely on our FOIL method.

- anonymous

so what will it look like after that is done?

- zepdrix

\[\large\rm =(5-x)^2\left[3\color{royalblue}{(5-x)(5-x)}+2(5-x)-1\right]\]FOIL'ing out the (5-x) brackets gives us:\[\large\rm =(5-x)^2\left[3\color{royalblue}{(25-10x+x^2)}+2(5-x)-1\right]\]

- zepdrix

Now we have to go a little further,
distributing the 3 to the blue stuff,
distributing the 2 to the next set of brackets.

- anonymous

ok ill try it and compare my answer with yours!:)

- anonymous

ok im ready!

- zepdrix

\[\large\rm =(5-x)^2\left[75-30x+3x^2+10-2x-1\right]\]So I guess we get something like this, ya?

- zepdrix

After that, we need to combine like-terms.

- anonymous

i got that also!

- zepdrix

yay!
now try to combine like-terms :D

- anonymous

I got (5-x)^2[84-32x+3x^2]

- zepdrix

Ok good.
Let's write it in standard form, with the x^2 coming first,
\[\large\rm =(5-x)^2\left[3x^2-32x+84\right]\]

- zepdrix

Hmm this step is going to be a bit tricky.
Do you remember `factor by grouping`?
That's what we need to apply here.

- anonymous

no I don't... can you show me what that would look like?

- zepdrix

When we're given a standard second degree polynomial,\[\large\rm ax^2+bx+c\]We multiply \(\large\rm a\) and \(\large\rm c\) and try to find factors of \(\large\rm ac\) that add to \(\large\rm b\).

- zepdrix

So for this problem...
We're multiplying \(\large\rm 3\) and \(\large\rm 84\)
and trying to find two factors of that number that add to \(\large\rm -32\)

- zepdrix

We don't want to deal with a big number like 3*84,
so instead we'll try a neat trick.
Let's break 84 into its prime factorization.

- zepdrix

\[\large\rm 3\cdot\color{orangered}{84}=3\cdot\color{orangered}{2\cdot42}=3\cdot\color{orangered}{2\cdot7\cdot6}\]Do you understand how I broke that down? :O

- anonymous

yes, now how did you know to stop breaking it down at 6?

- zepdrix

Ooo good question.
I could have gone further, I didn't actually succeed in getting all the primes lol.
All I was really trying to do was break it into "small" numbers,
so we can `test` some combinations.

- anonymous

WAIT

- zepdrix

But yes, let's break the 6 up, just in case it's necessary.\[\large\rm =2\cdot2\cdot3\cdot3\cdot7\]

- anonymous

is it (3x-28)(x-3) ?

- zepdrix

Sec, checking :)

- zepdrix

Hmmm no :d

- anonymous

why not? where did i go wrong?

- zepdrix

If you FOIL your result back out:
You get middle terms of: \(\large\rm -28x\) and \(\large\rm -9x\)
which unfortunately don't get us to \(\large\rm -32x\)
:(

- anonymous

wow a simple subraction mistake lol my b!

- anonymous

i got ahead of myself...

- zepdrix

hehe

- zepdrix

Let's try some combinations.
How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)

- zepdrix

So that gives us \(\large\rm 12\) and \(\large\rm 21\)
Hmm, no way to get \(\large\rm 32\) when combining those.
So we need to try another combination.

- zepdrix

Maybe this combination instead?
\(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)

- anonymous

no that didn't work, but would the combination of 6 and 14 work? with 14 being in the first group? (3x-14)(x-6)

- zepdrix

Woops! :)
Hold on, let's look at my last combination a sec.
I think it works.

- zepdrix

Given \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\),
So we have \(\large\rm 18\) and \(\large\rm 14\) ya?
Any way to combine these and get \(\large\rm 32\)?

- zepdrix

Oh you got to the right result,
you must have some sneaky shortcut method :P
I'm not sure what you did lol.

- anonymous

no lol just trial and erro method:P

- anonymous

but now what would it look like if i wrote it out completely?

- zepdrix

\(\large\rm 18+14=32\)
Which tells us that 18 and 14 are the factors that we want!
So we break up the -32x into -18x and -14x and do factor by grouping.\[\large\rm 3x^2\color{orangered}{-32x}+84\]\[\large\rm 3x^2\color{orangered}{-18x-14x}+84\]Factor a 3x out of each of the first two terms,\[\large\rm 3x(x-6)-14x+84\]Pull a -14 out of each of the other terms,\[\large\rm 3x(x-6)-14(x-6)\]Then pull an (x-6) out of everything,\[\large\rm (x-6)(3x-14)\]

- zepdrix

I know we're kind of past that step already,
but I just wanted to detail the `factor by grouping` method in case you needed to see it.

- anonymous

thank you! thats this is the explanation i was hoping for!

- zepdrix

So we've gotten to this point:\[\large\rm =(5-x)^2\left[\color{orangered}{3x^2-32x+84}\right]\]\[\large\rm =(5-x)^2\left[\color{orangered}{(3x-14)(x-6)}\right]\]We can drop the square brackets at this point as they're searching no purpose.\[\huge\rm (5-x)^2(3x-14)(x-6)\]And we're finally done!
yayyy team! \c:/

- zepdrix

serving# no purpose

- anonymous

aye!!!! thank you so much!

- zepdrix

typo :p

- zepdrix

np :3

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