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  • one year ago

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  1. anonymous
    • one year ago
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    This is a factoring trinomials problem. Tell me how you would usually solve this sort of problem?

  2. anonymous
    • one year ago
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    You can either: Factor By Grouping Complete The Square or Use the Quadratic Formula.

  3. anonymous
    • one year ago
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    But for this problem however, we'll be factoring by grouping. What is your understanding of this so far?

  4. anonymous
    • one year ago
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    Nothing. I missed a school day and this is in my packet. I tried looking online but I wasn't sure what this was called.

  5. anonymous
    • one year ago
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    Here are the steps to solve: (give me 2 mins)

  6. anonymous
    • one year ago
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    Okay.

  7. anonymous
    • one year ago
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    2.1 Factoring x^2-15x+50 The first term is, x^2 its coefficient is 1 . The middle term is, -15x its coefficient is -15 . The last term, "the constant", is +50 Step-1 : Multiply the coefficient of the first term by the constant 1 • 50 = 50 Step-2 : Find two factors of 50 whose sum equals the coefficient of the middle term, which is -15 . -50 + -1 = -51 -25 + -2 = -27 -10 + -5 = -15 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -10 and -5 x2 - 10x - 5x - 50 Step-4 : Add up the first 2 terms, pulling out like factors : x • (x-10) Add up the last 2 terms, pulling out common factors : 5 • (x-10) Step-5 : Add up the four terms of step 4 : (x-5) • (x-10) Which is the desired factorization Final result : (x - 5) • (x - 10)

  8. anonymous
    • one year ago
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    You can always check your work by distributing/FOILing

  9. anonymous
    • one year ago
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    Also, if you need more help on the basics of this, please go to:

  10. anonymous
    • one year ago
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    Khan academy is an awesome reliable source to help you out. Any Q's? :)

  11. anonymous
    • one year ago
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    I'm just gonna have to read through all of that to make sure I understand it. I really appreciate you writing all of that out and for the website too. :)

  12. anonymous
    • one year ago
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    No problem! Tag me if you need more help with factoring polynomials! (It's my specialty haha)

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