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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    split the middle term, do you know how to do that?

  2. anonymous
    • one year ago
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    kind of but could you please walk me through the steps

  3. anonymous
    • one year ago
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    I'll try when you split the middle term of a quadratic polynomial \[p(x)=ax^2+bx+c\] you are looking to split b into two numbers say, f and g such that \[b=f+g\] and \[f \times g =a \times c\] Do you understand this?

  4. anonymous
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    Ok now calculate \[a \times c\] for the polynomial u r given

  6. anonymous
    • one year ago
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    |dw:1438524585865:dw|

  7. anonymous
    • one year ago
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    okay keep going i understand

  8. anonymous
    • one year ago
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    can u find the product of a and c for me

  9. anonymous
    • one year ago
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    -12

  10. anonymous
    • one year ago
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    ok good, now what is your b?

  11. anonymous
    • one year ago
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    -4

  12. anonymous
    • one year ago
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    great, now can you think of two numbers u can split b(that is -4 in this case) into, such that the product of those numbers will equal -12 and adding those numbers will give u back -4

  13. anonymous
    • one year ago
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    -6 and 2

  14. anonymous
    • one year ago
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    good job!! now just split b and work it out as \[12x^2+(-6+2)x-1=12x^2-6x+2x-1\]

  15. anonymous
    • one year ago
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    factorize first 2 terms together and last 2 terms together

  16. anonymous
    • one year ago
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    what two terms

  17. anonymous
    • one year ago
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    |dw:1438524844892:dw|

  18. anonymous
    • one year ago
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    take out as much factors out as u can from the first 2 terms

  19. anonymous
    • one year ago
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    |dw:1438524921185:dw| What does that say?

  20. anonymous
    • one year ago
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    it's 6x, for the first 2 terms what can u take out??what is common in 12x^2 and -6x ??

  21. anonymous
    • one year ago
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    6x(2x-1)

  22. anonymous
    • one year ago
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    good, now so now we have \[6x(2x-1)+(2x-1)\] Now we have two terms, you can take 2x-1 out, what are you left with?

  23. anonymous
    • one year ago
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    6x

  24. anonymous
    • one year ago
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    nope, you are missing something!! \[(2x-1)(6x+??)\]

  25. anonymous
    • one year ago
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    1?

  26. anonymous
    • one year ago
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    yep! and there you have your factors, nice and clean \[(2x-1)(6x+1)\] Just to verify, let's open these up \[(2x \times 6x)+( -1 \times 6x)+(2x \times 1)+(-1 \times 1)=12x^2-6x+2x-1=12x^2-4x-1\]

  27. anonymous
    • one year ago
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    12x^2-4x-1

  28. anonymous
    • one year ago
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    Thanks

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