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BloomLocke367

  • one year ago

Help with factoring?

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  1. BloomLocke367
    • one year ago
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    \(\Huge x^2-10x+25-36y^2\)

  2. BloomLocke367
    • one year ago
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    at first I thought about factoring by grouping, but 25 and -36 don't have a GCF. So, I'm not sure what to do.

  3. BloomLocke367
    • one year ago
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    I'm thinking it may be prime... but I'm not sure.

  4. BloomLocke367
    • one year ago
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    @ganeshie8

  5. BloomLocke367
    • one year ago
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    @phi

  6. anonymous
    • one year ago
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    Think of it like\[\left( x^2- 10x+25\right)-y^2\]Can you factor the trinomial in the brackets?

  7. BloomLocke367
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    *36y^2 Sorry

  9. anonymous
    • one year ago
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    OK. Go ahead. What do you get?

  10. BloomLocke367
    • one year ago
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    \((x-5)(x-5)\) or\( (x-5)^2\)

  11. anonymous
    • one year ago
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    Excellent. So now you have\[\left( x-5 \right)^2 - 36y^2\]This is a difference of squares. Do you know how to factor these?

  12. BloomLocke367
    • one year ago
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    I didddd, but it's been like a year since I have. I forgot what to use. :/

  13. BloomLocke367
    • one year ago
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    I should have kept my notes from Algebra 2 XD

  14. anonymous
    • one year ago
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    OK. In general, factoring a difference of squares looks like\[a^2-b^2 = \left( a+b \right)\left( a-b \right)\]In your question, you have \(a=x-5\) and \(b=6y\). Can you put it all together now?

  15. BloomLocke367
    • one year ago
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    yes, thank you. I'm going to write that down XD

  16. anonymous
    • one year ago
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    You're welcome.

  17. BloomLocke367
    • one year ago
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    So I have \((x-5)^2-(6y)^2=(x-5+6y)(x-5-6y)\)

  18. BloomLocke367
    • one year ago
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    @ospreytriple

  19. anonymous
    • one year ago
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    Correct. Well done.

  20. BloomLocke367
    • one year ago
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    I'm not done yet, am I?

  21. anonymous
    • one year ago
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    Can you simplify these factors any further?

  22. BloomLocke367
    • one year ago
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    ummmmm, I don't think so

  23. BloomLocke367
    • one year ago
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    @ospreytriple

  24. anonymous
    • one year ago
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    Good. Then you're finished. Sorry, I am having connectivity issues.

  25. BloomLocke367
    • one year ago
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    and while you're here, can you tell me how to find the difference of two cubes? and it's alright :)

  26. anonymous
    • one year ago
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    Difference of cubes: \[a^3-b^3 = \left( a-b \right)\left( a^2+ab+b^2 \right)\]

  27. BloomLocke367
    • one year ago
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    ok, thanks!

  28. anonymous
    • one year ago
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    Oh, and sum of cubes:\[a^3 + b^3 = \left( a+b \right)\left( a^2-ab+b^2 \right)\]

  29. anonymous
    • one year ago
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    NP.

  30. BloomLocke367
    • one year ago
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    thank youuuuuuuuuuuuuu

  31. BloomLocke367
    • one year ago
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    also, is this prime? \(x^2+36\)

  32. anonymous
    • one year ago
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    Not sure if "prime" is the correct term, but it is unfactorable with real numbers.

  33. BloomLocke367
    • one year ago
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    prime and unfactorable mean the same thing when it comes to factoring.

  34. anonymous
    • one year ago
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    Fair enough. I've never come across that term used to describe polynomials.

  35. BloomLocke367
    • one year ago
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    haha, yeah. and for \(x^2-27\) I got \(x^3-3^3=(x-3)(x^2+3x+9)\). Is that correct?

  36. BloomLocke367
    • one year ago
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    @Preetha is that correct? ^^

  37. Nnesha
    • one year ago
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    :P

  38. BloomLocke367
    • one year ago
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    you're still here XD is that right?

  39. Nnesha
    • one year ago
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    lol do you mean \[\rm x^3-27 \] ?

  40. BloomLocke367
    • one year ago
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    yesss, my bad XD

  41. BloomLocke367
    • one year ago
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    I'm pretty sure it's correct because when you distribute it back out, you get \(x^3-27\)

  42. Nnesha
    • one year ago
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    yes right :) i know you already know but just want to share SOAP for signs |dw:1440020416579:dw|

  43. BloomLocke367
    • one year ago
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    soooo I did get it

  44. Nnesha
    • one year ago
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    :)

  45. BloomLocke367
    • one year ago
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    okay, I have one more.

  46. Nnesha
    • one year ago
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    ok:)

  47. anonymous
    • one year ago
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    Sorry. Lost connection again.

  48. BloomLocke367
    • one year ago
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    \(6x^2-7xy-5y^2\), I'm not sure what to do... does \(5y^2\) count as c? Because I know that when \(a\ne 1\) you take \(ac\) and find the factors of those that multiply to get c but add to get b.

  49. anonymous
    • one year ago
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    That's right. What do you think? 2 number that multiply to give -30 and add to -7

  50. Nnesha
    • one year ago
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    \[a\cancel{=}1\] i'm not agree if a =1 you can multiply ac

  51. BloomLocke367
    • one year ago
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    I know. but you could just use c XD

  52. BloomLocke367
    • one year ago
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    and -10 and 3

  53. Nnesha
    • one year ago
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    and when \[a\cancel{=}1\] when you multiply 2 numbers you should get product of AC not just C

  54. BloomLocke367
    • one year ago
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    I knowwwwwwww. XD lol I just didn't know if -5y^2 was C or not.

  55. Nnesha
    • one year ago
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    LOL just saying :P

  56. BloomLocke367
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @BloomLocke367 \(6x^2-7xy-5y^2\), I'm not sure what to do... does \(5y^2\) count as c? Because I know that when \(a\ne 1\) you take \(ac\) and find the factors of those that multiply to get c but add to get b. \(\color{#0cbb34}{\text{End of Quote}}\) I meant to say that multiply to get ac.. now I see why you corrected me @Nnesha XD

  57. anonymous
    • one year ago
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    Perfect. So rewrite as\[6x^2+3xy-10xy-5y^2\]Regroup as two binomials\[\left( 6x^2+3xy \right)-\left( 10xy+5y^2 \right)\]Factor common factor out of each binomial.

  58. BloomLocke367
    • one year ago
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    ohhh okay I got it, thanks! I gotta go, bye! :) thanks so so so so much for the help!

  59. anonymous
    • one year ago
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    You're welcome

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