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Math Help - Factoring and Simplification, Very Basic

  1. #1
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    Factoring and Simplification, Very Basic

    Hi all,
    First post as you can see. I'm doing high school maths and really need to improve. I can do every exercise in the basic algebra chapter except for these 2. I have no idea why, they've just stumped me.

    So any help would be much appreciated.




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  2. #2
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    Quote Originally Posted by lucius View Post
    Hi all,
    First post as you can see. I'm doing high school maths and really need to improve. I can do every exercise in the basic algebra chapter except for these 2. I have no idea why, they've just stumped me.

    So any help would be much appreciated.




    For the first one, note that 2m^2+3mn+n^2=2m^2+(2mn+mn)+n^2

    For the second one, note that c^2-4=(c+2)(c-2)
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  3. #3
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    That was very helpful Krizalid I had them done in no time. Thanks!
    Last edited by lucius; June 12th 2007 at 07:20 PM. Reason: Problems Solved
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  4. #4
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    Ugh

    Ok I've run into two more.




    Thanks for the help on the last one, much appreciated.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lucius View Post
    Ok I've run into two more.




    Thanks for the help on the last one, much appreciated.
    \frac {8q^2 + 2qr - 6r^2 + 8qs - 6rs}{8q^2 - 2qr - 3r^2 + 12 qs - 9rs}

    = \frac {2 \left( 4q^2 + qr - 3r^2 + 4qs - 3rs \right)}{(2q + r)(4q - 3r) + 3s(4q - 3r)}

    = \frac {2 [(4q - 3r)(q + r) + s(4q - 3r)]}{(4q - 3r)(2q + r + 3s)}

    = \frac {2(4q - 3r)(q + r + s)}{(4q - 3r)(2q + r + 3s)}

    = \frac {2(q + r + s)}{2q + r + 3s}




    \frac {15g^2 k^3 + 35g^3 k^2 - 30g^4 k}{8gk^4 + 44g^2 k^3 + 60g^3 k^2}

    = \frac {5g^2 k \left( 3k^2 + 7gk - 6g^2 \right)}{4g k^2 \left(  2k^2 + 11gk + 15g^2\right)}

    = \frac {5g^2 k (3k - 2g)(k + 3g)}{4g k^2 (2k + 5g)(k + 3g)}

    = \frac {5g(3k - 2g)}{4k(2k + 5g)}
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  6. #6
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    Quote Originally Posted by lucius View Post
    Ok I've run into two more.




    Thanks for the help on the last one, much appreciated.
    Hello,

    to #1:

    You have to use the method here that you add a sum with the value zero to the numerator and to the denominator. I'll put the two summands in question into boxes:

    \frac{8q^2+2qr + \boxed{6qr} - \boxed{6qr}-6r^2+8qs - 6rs}{8q^2-2qr -\boxed{4qr}+\boxed{4qr}-3r^2+12qs-9rs} = \text{     } Now re-arrange numerator and denomintor:

    \frac{8q^2+8qr + 8qs-6qr-6r^2 - 6rs}{8q^2-6qr +4qr-3r^2+12qs-9rs} =\ \ \ Now factorize numerator and denominator:

    \frac{8q(q+r + s)-6r(q+r+s)}{2q(4q-3r) +r(4q-3r)+3s(4q-3r)} =\ \ \ Now factor again and cancel the common factor which is: (4q - 3r)

    I'll leave the rest for you.
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  7. #7
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    You've all been so helpful! Thanks a lot. Until nex time.
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  8. #8
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    Hello, lucius!

    Welcome aboard!


    Factor: . 2m^2 + 3mn + n^2
    There may be two reasons for your difficulty:
    . . [1] There are two variables
    . . [2] That "2" in front.

    Just as we "split" the first term into two factors,
    . . we can do the same with the last term.

    It looks like this: . (\:m\qquad n)(\:m \qquad n)

    Since 2m^2 factors into (2m)(m) .and n^2 factors into (n)(n),
    . . we have: . (2m\qquad n)(m\qquad n)

    Now insert the signs that make the "middle term" come out to 3mn.

    Answer: . (2m + n)(m + n)



    Simplify: . \frac{1}{c+2} + \frac{2}{c-2} + \frac{4}{c^2-4}
    I'll take baby-steps . . .

    Factoring the denominators, we have: . \frac{1}{c+2} + \frac{2}{c-2} + \frac{4}{(c+2)(c-2)}

    To add or subtract fractions, we must have a common denominator.
    . . In this problem, it is: (c+2)(c-2)

    Now we must "convert" each fraction so it has (c+2)(c-2) in the denominator.

    Multiply the first fraction by . \frac{c-2}{c-2}
    Multiply the second fraction by . \frac{c+2}{c+2}

    The problem looks like this: . \frac{c-2}{c-2}\cdot\frac{1}{c+2} \:+\: \frac{c+2}{c+2}\cdot\frac{2}{c-2} \:+\: \frac{4}{(c+2)(c-2)}

    . . and we have: . \frac{c-2}{(c+2)(c-2)} + \frac{2(c+2)}{(c+2)(c-2)} + \frac{4}{(c+2)(c-2)}


    The fractions have a common denominator, so we can combine them.
    . . Combine the numerators and "keep" the common denominator.

    So we have: . \frac{c - 2 + 2(c + 2) + 4}{(c+2)(c-2)} \;=\;\frac{c - 2 + 2c + 4 + 4}{(c+2)(c-2)} \;=\;\frac{3c+6}{(c+2)(c-2)}


    Are we done? . . . Well, no. .The fraction can be reduced.

    Factor: . \frac{3(c + 2)}{(c + 2)(c-2)}

    And cancel the (c+2)\;\; \cdots \;\;\text{Answer: }\;\frac{3}{c-2}

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  9. #9
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    Well, after an "of course!" moment or two I'm right with everything now.

    Thanks a lot everyone, such fast and helpful responses! I'm very grateful.
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