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

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  1. zepdrix
    • one year ago
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    Hey :) \[\large\rm 3(5-x)^4+2(5-x)^3-(5-x)^2\]

  2. zepdrix
    • one year ago
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    \[\large\rm 3(5-x)(5-x)(5-x)^2+2(5-x)(5-x)^2-(5-x)^2\]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
    • one year ago
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    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
    • one year ago
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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.

  5. anonymous
    • one year ago
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    why did you keep (5-x)^2 at the end? why not just write it out another time?

  6. anonymous
    • one year ago
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    is that a GCF?

  7. zepdrix
    • one year ago
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    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
    • one year ago
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    okaayyyy i got it, now would the -? at the end be -1?

  9. zepdrix
    • one year ago
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    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
    • one year ago
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    okay so now what? can it be simplified further?

  11. zepdrix
    • one year ago
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    \[\large\rm =(5-x)^2\left[\color{royalblue}{3(5-x)(5-x)+2(5-x)-1}\right]\] `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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    so what will it look like after that is done?

  16. zepdrix
    • one year ago
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    \[\large\rm =(5-x)^2\left[3\color{royalblue}{(5-x)(5-x)}+2(5-x)-1\right]\]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
    • one year ago
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    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
    • one year ago
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    ok ill try it and compare my answer with yours!:)

  19. anonymous
    • one year ago
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    ok im ready!

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

  21. zepdrix
    • one year ago
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    After that, we need to combine like-terms.

  22. anonymous
    • one year ago
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    i got that also!

  23. zepdrix
    • one year ago
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    yay! now try to combine like-terms :D

  24. anonymous
    • one year ago
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    I got (5-x)^2[84-32x+3x^2]

  25. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    no I don't... can you show me what that would look like?

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

  32. anonymous
    • one year ago
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    yes, now how did you know to stop breaking it down at 6?

  33. zepdrix
    • one year ago
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    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
    • one year ago
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    WAIT

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

  36. anonymous
    • one year ago
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    is it (3x-28)(x-3) ?

  37. zepdrix
    • one year ago
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    Sec, checking :)

  38. zepdrix
    • one year ago
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    Hmmm no :d

  39. anonymous
    • one year ago
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    why not? where did i go wrong?

  40. zepdrix
    • one year ago
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    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
    • one year ago
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    wow a simple subraction mistake lol my b!

  42. anonymous
    • one year ago
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    i got ahead of myself...

  43. zepdrix
    • one year ago
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    hehe

  44. zepdrix
    • one year ago
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    Let's try some combinations. How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)

  45. zepdrix
    • one year ago
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    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
    • one year ago
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    Maybe this combination instead? \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)

  47. anonymous
    • one year ago
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    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
    • one year ago
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    Woops! :) Hold on, let's look at my last combination a sec. I think it works.

  49. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    no lol just trial and erro method:P

  52. anonymous
    • one year ago
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    but now what would it look like if i wrote it out completely?

  53. zepdrix
    • one year ago
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    \(\large\rm 18+14=32\) 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
    • one year ago
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    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
    • one year ago
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    thank you! thats this is the explanation i was hoping for!

  56. zepdrix
    • one year ago
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    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
    • one year ago
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    serving# no purpose

  58. anonymous
    • one year ago
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    aye!!!! thank you so much!

  59. zepdrix
    • one year ago
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    typo :p

  60. zepdrix
    • one year ago
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    np :3

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