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Math Help - Need urgent help on factorising!

  1. #1
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    Need urgent help on factorising!

    Just a few questions I am having trouble with here before my exam:
    Factorize
    (1)
    p^6-64
    (2) 25e^2-30ef-16f^2
    (3) 2a^2-2b^2+5a+2am-5b-2bm

    Thanks for the help.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    1.  p^{6} - 64 = (p-2)(p+2)(p^{2}+2p+4)(p^{2}-2p+4) .

    Try the rest. For the first one I used the rational roots theorem and synthetic division to find the roots. Then used polynomial long division.

    So:  \frac{\pm 1,2,4,8,16,32,64}{\pm 1} are canidates for possible rational roots.


    2. For this one, it is just like factoring a regular quadratic equation like  x^{2} + 2x + 1 .


    3. I am not sure that this can be factored.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by holmesb View Post

    (1)
    p^6-64
    We can be a bit more analytic than using the rational roots theorem. There are two ways of looking at this one:
    First:
    p^4 - 64 = (p^3)^2 - 8^2

    = (p^3 +8)(p^3 - 8)

    = (p^3 + 2^3)(p^3 - 2^3)

    Now,
    a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

    So
    p^3 + 2^3 = (p + 2)(p^2 - 2p + 4)
    p^3 - 2^3 = (p - 2)(p^2 + 2p + 4)

    Thus
    p^4 - 64 = (p + 2)(p^2 - 2p + 4)(p - 2)(p^2 + 2p + 4)

    Second:
    p^6 - 64 = (p^2)^3 - 4^3

    = (p^2 - 4)(p^4 + 4p^2 + 16)

    = (p + 2)(p - 2)(p^4 + 4p^2 + 16)

    Now, p^4 + 4p^2 + 16 is not particularly easy to factor, but note that:
    p^4 + 4p^2 + 16 = (p^4 + 4p^2 + 16) + (4p^2 - 4p^2) (Just adding 0 to the expression)

    = (p^4 + 8p^2 + 16) - 4p^2

    = (p^2 + 4)^2 - (2p)^2 <-- The difference of two squares

    = ([p^2 + 4] + 2p)([p^2 + 4) - 2p)

    = (p^2 + 2p + 4)(p^2 - 2p + 4)

    So again we get:
    p^6 - 64 = (p + 2)(p - 2)(p^2 + 2p + 4)(p^2 - 2p + 4)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by holmesb View Post
    (3) 2a^2-2b^2+5a+2am-5b-2bm
    We can try a variety of things here:
    One way is:
    2a^2 - 2b^2 + 5a + 2am - 5b - 2bm

    = (2a^2 + 5a) - (2b^2 + 5b) + 2am - 2bm

     = 2 \left ( a^2 + \frac{5}{2}a \right ) - 2 \left ( b^2 + \frac{5}{2}b \right ) + 2am - 2bm

    Now complete the square on the first two terms. We have to add \left ( \frac{5}{4} \right ) ^2 = \frac{25}{16} to each term so:
    =  2 \left ( a^2 + \frac{5}{2}a  + \frac{25}{16} \right ) - 2 \left ( b^2 + \frac{5}{2}b + \frac{25}{16} \right ) + 2am - 2bm - 2 \cdot 2 \cdot \frac{25}{16}
    (The second factor of 2 comes from the fact that each of the first two factors is multiplied by 2. If you doubt me, expand it all out. You'll see that I'm just adding a total of 0 to the expression.)

    = \left ( a + \frac{5}{4} \right ) ^2 - \left ( b + \frac{5}{4} \right ) ^2 + 2am - 2bm - \frac{25}{4}

     = \left ( a + \frac{5}{4} \right ) ^2 - \left ( b + \frac{5}{4} \right ) ^2 + 2m(a - b) - \frac{25}{4}

    Another way is:
    2a^2 - 2b^2 + 5a + 2am - 5b - 2bm

    (2a^2 + 2am) - (2b^2 + 2bm) + 5a - 5b

    2(a^2 + am) - 2(b^2 + bm) + 5a - 5b<-- Again complete the square:

    = 2 \left ( a^2 + am + \frac{m^2}{4} \right ) - 2 \left ( b + bm + \frac{m^2}{4} \right ) + 5a - 5b - 2 \cdot 2 \cdot \frac{m^2}{4}

    = 2 \left ( a + \frac{m}{2} \right ) ^2 - 2 \left ( b + \frac{m}{2} \right ) ^2 + 5(a - b) - m^2

    -Dan
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