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

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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

Mathematics
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Hey :) \[\large\rm 3(5-x)^4+2(5-x)^3-(5-x)^2\]
\[\large\rm 3(5-x)(5-x)(5-x)^2+2(5-x)(5-x)^2-(5-x)^2\]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: \(\large\rm x^4=x\cdot x\cdot x^2\))
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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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.
why did you keep (5-x)^2 at the end? why not just write it out another time?
is that a GCF?
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 :)
okaayyyy i got it, now would the -? at the end be -1?
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.\[\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-1\right]\]
okay so now what? can it be simplified further?
\[\large\rm =(5-x)^2\left[\color{royalblue}{3(5-x)(5-x)+2(5-x)-1}\right]\] `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.
Do you understand how to multiply (5-x)(5-x)? :) Think you can get the blue stuff into the form ax^2+bx+c?
I understand that they are binomials and can be simplified by using FOIL correct? or do we not use this equation for this?
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.
so what will it look like after that is done?
\[\large\rm =(5-x)^2\left[3\color{royalblue}{(5-x)(5-x)}+2(5-x)-1\right]\]FOIL'ing out the (5-x) brackets gives us:\[\large\rm =(5-x)^2\left[3\color{royalblue}{(25-10x+x^2)}+2(5-x)-1\right]\]
Now we have to go a little further, distributing the 3 to the blue stuff, distributing the 2 to the next set of brackets.
ok ill try it and compare my answer with yours!:)
ok im ready!
\[\large\rm =(5-x)^2\left[75-30x+3x^2+10-2x-1\right]\]So I guess we get something like this, ya?
After that, we need to combine like-terms.
i got that also!
yay! now try to combine like-terms :D
I got (5-x)^2[84-32x+3x^2]
Ok good. Let's write it in standard form, with the x^2 coming first, \[\large\rm =(5-x)^2\left[3x^2-32x+84\right]\]
Hmm this step is going to be a bit tricky. Do you remember `factor by grouping`? That's what we need to apply here.
no I don't... can you show me what that would look like?
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 \(\large\rm b\).
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 \(\large\rm -32\)
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.
\[\large\rm 3\cdot\color{orangered}{84}=3\cdot\color{orangered}{2\cdot42}=3\cdot\color{orangered}{2\cdot7\cdot6}\]Do you understand how I broke that down? :O
yes, now how did you know to stop breaking it down at 6?
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.
WAIT
But yes, let's break the 6 up, just in case it's necessary.\[\large\rm =2\cdot2\cdot3\cdot3\cdot7\]
is it (3x-28)(x-3) ?
Sec, checking :)
Hmmm no :d
why not? where did i go wrong?
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 \(\large\rm -32x\) :(
wow a simple subraction mistake lol my b!
i got ahead of myself...
hehe
Let's try some combinations. How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)
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.
Maybe this combination instead? \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)
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)
Woops! :) Hold on, let's look at my last combination a sec. I think it works.
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 \(\large\rm 32\)?
Oh you got to the right result, you must have some sneaky shortcut method :P I'm not sure what you did lol.
no lol just trial and erro method:P
but now what would it look like if i wrote it out completely?
\(\large\rm 18+14=32\) 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,\[\large\rm (x-6)(3x-14)\]
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.
thank you! thats this is the explanation i was hoping for!
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:/
serving# no purpose
aye!!!! thank you so much!
typo :p
np :3

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