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Math Help - Last two, I promise (for tonite)

  1. #1
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    Last two, I promise (for tonite)

    I'm not sure if I know how to write these out, but here goes: (multiply and simplify)

    k^2+11k+28 X k^2+8k
    k^2+12k+32 X k^2+4k-21

    ADD and simplify:
    m^2-8m + 15
    m-5 m-5

    That is m^2-8m over m-5 PLUS
    15 over m-5
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  2. #2
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    The first one (k^2+11k+28)(k^2+8k)=k^4+8k^3+11k^3+88k^2+28k^2+22 4k=k^4+19k^3+116k^2+224k

    That looks like it to me unless I messed up my mental math. Do you see the method?
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  3. #3
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    Quote Originally Posted by Jameson View Post
    The first one (k^2+11k+28)(k^2+8k)=k^4+8k^3+11k^3+88k^2+28k^2+22 4k=k^4+19k^3+116k^2+224k

    That looks like it to me unless I messed up my mental math. Do you see the method?
    Not really, but I'm a bit dense tonite. My answer was nowhere near that.

    I think I will go beddy-bye. It is 12:03 a.m. here. My eyes are soooo tired

    Thanks. Bye.
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  4. #4
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    Quote Originally Posted by Suwanee View Post
    ADD and simplify:
    m^2-8m + 15
    m-5 m-5
    Both fractions have the same denominator, so you can just add the numerators:
    (m^2 - 8m + 15)/(m - 5)

    Now, the numerator factors: m^2 - 8m + 15 = (m-5)(m-3), so we get
    [(m-5)(m-3)]/(m-5) = m - 3 because the m-5 terms cancel.

    -Dan
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