## 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(5-x)^4+2(5-x)^3-(5-x)^2

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1. zepdrix

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

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: $$\large\rm x^4=x\cdot x\cdot x^2$$)

3. zepdrix

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

4. 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.

5. anonymous

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

6. anonymous

is that a GCF?

7. 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 :)

8. anonymous

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

9. 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.$\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-1\right]$

10. anonymous

okay so now what? can it be simplified further?

11. 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.

12. 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?

13. anonymous

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

14. 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.

15. anonymous

so what will it look like after that is done?

16. 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:$\large\rm =(5-x)^2\left[3\color{royalblue}{(25-10x+x^2)}+2(5-x)-1\right]$

17. 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.

18. anonymous

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

19. anonymous

20. zepdrix

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

21. zepdrix

After that, we need to combine like-terms.

22. anonymous

i got that also!

23. zepdrix

yay! now try to combine like-terms :D

24. anonymous

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

25. zepdrix

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

26. 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.

27. anonymous

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

28. 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 $$\large\rm b$$.

29. 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 $$\large\rm -32$$

30. 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.

31. 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

32. anonymous

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

33. 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.

34. anonymous

WAIT

35. zepdrix

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

36. anonymous

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

37. zepdrix

Sec, checking :)

38. zepdrix

Hmmm no :d

39. anonymous

why not? where did i go wrong?

40. 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 $$\large\rm -32x$$ :(

41. anonymous

wow a simple subraction mistake lol my b!

42. anonymous

43. zepdrix

hehe

44. zepdrix

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

45. 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.

46. zepdrix

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

47. 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)

48. zepdrix

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

49. 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 $$\large\rm 32$$?

50. zepdrix

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

51. anonymous

no lol just trial and erro method:P

52. anonymous

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

53. 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)$

54. 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.

55. anonymous

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

56. 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:/

57. zepdrix

serving# no purpose

58. anonymous

aye!!!! thank you so much!

59. zepdrix

typo :p

60. zepdrix

np :3