anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zepdrix
  • zepdrix
Hey :) \[\large\rm 3(5-x)^4+2(5-x)^3-(5-x)^2\]
zepdrix
  • 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
  • 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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zepdrix
  • 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
  • anonymous
why did you keep (5-x)^2 at the end? why not just write it out another time?
anonymous
  • anonymous
is that a GCF?
zepdrix
  • 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
  • anonymous
okaayyyy i got it, now would the -? at the end be -1?
zepdrix
  • 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
  • anonymous
okay so now what? can it be simplified further?
zepdrix
  • 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
  • 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
  • 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
  • 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
  • anonymous
so what will it look like after that is done?
zepdrix
  • 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
  • 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
  • anonymous
ok ill try it and compare my answer with yours!:)
anonymous
  • anonymous
ok im ready!
zepdrix
  • 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
  • zepdrix
After that, we need to combine like-terms.
anonymous
  • anonymous
i got that also!
zepdrix
  • zepdrix
yay! now try to combine like-terms :D
anonymous
  • anonymous
I got (5-x)^2[84-32x+3x^2]
zepdrix
  • 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
  • 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
  • anonymous
no I don't... can you show me what that would look like?
zepdrix
  • 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
  • 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
  • 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
  • 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
  • anonymous
yes, now how did you know to stop breaking it down at 6?
zepdrix
  • 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
  • anonymous
WAIT
zepdrix
  • zepdrix
But yes, let's break the 6 up, just in case it's necessary.\[\large\rm =2\cdot2\cdot3\cdot3\cdot7\]
anonymous
  • anonymous
is it (3x-28)(x-3) ?
zepdrix
  • zepdrix
Sec, checking :)
zepdrix
  • zepdrix
Hmmm no :d
anonymous
  • anonymous
why not? where did i go wrong?
zepdrix
  • 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
  • anonymous
wow a simple subraction mistake lol my b!
anonymous
  • anonymous
i got ahead of myself...
zepdrix
  • zepdrix
hehe
zepdrix
  • zepdrix
Let's try some combinations. How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)
zepdrix
  • 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
  • zepdrix
Maybe this combination instead? \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)
anonymous
  • 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
  • zepdrix
Woops! :) Hold on, let's look at my last combination a sec. I think it works.
zepdrix
  • 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
  • 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
  • anonymous
no lol just trial and erro method:P
anonymous
  • anonymous
but now what would it look like if i wrote it out completely?
zepdrix
  • 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
  • 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
  • anonymous
thank you! thats this is the explanation i was hoping for!
zepdrix
  • 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
  • zepdrix
serving# no purpose
anonymous
  • anonymous
aye!!!! thank you so much!
zepdrix
  • zepdrix
typo :p
zepdrix
  • zepdrix
np :3

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