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

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hey :) \[\large\rm 3(5x)^4+2(5x)^3(5x)^2\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 3(5x)(5x)(5x)^2+2(5x)(5x)^2(5x)^2\]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
 one year ago
Best ResponseYou've already chosen the best response.1From here we can clearly see what they have in common:\[\large\rm 3(5x)(5x)\color{#DD4747}{(5x)^2}+2(5x)\color{#DD4747}{(5x)^2}\color{#DD4747}{(5x)^2}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh let's add the brackets on the outside of everything, maybe that will make things a little more clear.\[\large\rm \left[3(5x)(5x)\color{#DD4747}{(5x)^2}+2(5x)\color{#DD4747}{(5x)^2}\color{#DD4747}{(5x)^2}\right]\]So we're doing is, we're pulling this common factor out of the square brackets.\[\large\rm =\color{#DD4747}{(5x)^2}\left[3(5x)(5x)+2(5x)?\right]\]Gotta be careful here. What am I going to get in my last term? :) When you fully factor something out of itself.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why did you keep (5x)^2 at the end? why not just write it out another time?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Yes. It appears that (5x)^2 was the GCF! So we didn't really want to break it down into (5x)(5x), you could though :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okaayyyy i got it, now would the ? at the end be 1?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good! When you divided (5x)^2 out of (5x)^2, you're left with a 1. You're not left with a 0, I just wanted to make sure that was clear.\[\large\rm =\color{#DD4747}{(5x)^2}\left[3(5x)(5x)+2(5x)1\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so now what? can it be simplified further?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm =(5x)^2\left[\color{royalblue}{3(5x)(5x)+2(5x)1}\right]\] `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
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand how to multiply (5x)(5x)? :) Think you can get the blue stuff into the form ax^2+bx+c?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand that they are binomials and can be simplified by using FOIL correct? or do we not use this equation for this?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good :) 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
 one year ago
Best ResponseYou've already chosen the best response.0so what will it look like after that is done?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm =(5x)^2\left[3\color{royalblue}{(5x)(5x)}+2(5x)1\right]\]FOIL'ing out the (5x) brackets gives us:\[\large\rm =(5x)^2\left[3\color{royalblue}{(2510x+x^2)}+2(5x)1\right]\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Now we have to go a little further, distributing the 3 to the blue stuff, distributing the 2 to the next set of brackets.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok ill try it and compare my answer with yours!:)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm =(5x)^2\left[7530x+3x^2+102x1\right]\]So I guess we get something like this, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1After that, we need to combine liketerms.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1yay! now try to combine liketerms :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got (5x)^2[8432x+3x^2]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ok good. Let's write it in standard form, with the x^2 coming first, \[\large\rm =(5x)^2\left[3x^232x+84\right]\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm this step is going to be a bit tricky. Do you remember `factor by grouping`? That's what we need to apply here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no I don't... can you show me what that would look like?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1When 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1We 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm 3\cdot\color{orangered}{84}=3\cdot\color{orangered}{2\cdot42}=3\cdot\color{orangered}{2\cdot7\cdot6}\]Do you understand how I broke that down? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, now how did you know to stop breaking it down at 6?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ooo 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1But yes, let's break the 6 up, just in case it's necessary.\[\large\rm =2\cdot2\cdot3\cdot3\cdot7\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it (3x28)(x3) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why not? where did i go wrong?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.0wow a simple subraction mistake lol my b!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got ahead of myself...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's try some combinations. How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1Maybe this combination instead? \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no that didn't work, but would the combination of 6 and 14 work? with 14 being in the first group? (3x14)(x6)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Woops! :) Hold on, let's look at my last combination a sec. I think it works.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Given \(\large\rm 2\cdot3\cdot3\) 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
 one year ago
Best ResponseYou've already chosen the best response.1Oh you got to the right result, you must have some sneaky shortcut method :P I'm not sure what you did lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no lol just trial and erro method:P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but now what would it look like if i wrote it out completely?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\rm 18+14=32\) 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.\[\large\rm 3x^2\color{orangered}{32x}+84\]\[\large\rm 3x^2\color{orangered}{18x14x}+84\]Factor a 3x out of each of the first two terms,\[\large\rm 3x(x6)14x+84\]Pull a 14 out of each of the other terms,\[\large\rm 3x(x6)14(x6)\]Then pull an (x6) out of everything,\[\large\rm (x6)(3x14)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0thank you! thats this is the explanation i was hoping for!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So we've gotten to this point:\[\large\rm =(5x)^2\left[\color{orangered}{3x^232x+84}\right]\]\[\large\rm =(5x)^2\left[\color{orangered}{(3x14)(x6)}\right]\]We can drop the square brackets at this point as they're searching no purpose.\[\huge\rm (5x)^2(3x14)(x6)\]And we're finally done! yayyy team! \c:/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0aye!!!! thank you so much!
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