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Math Help - Factoring?

  1. #1
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    Factoring?

    1) x^2 - x - a - a^2

    2) x^4 - 3x^2 - 4

    3) x^4 - 8x^2 + 16

    4) xy + 2 - 2x - y

    5) 5s^2 - 16s + 12

    6) 6a^3 - 7a^2b - 2ab^2

    How do I factor these completely? I'm really lost right now
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  2. #2
    Eater of Worlds
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    Practice is what factoring requires.

    3) x^{4} - 8x^{2} + 16
    What two numbers when multiplied equal 16 and when added equal -8

    How about -4 and -4

    x^{4}-4x^{2}-4x^{2}+16

    x^{2}(x^{2}-4)-4(x^{2}-4)

    Notice the difference of two squares?

    (x-2)(x+2)(x-2)(x+2)=(x+2)^{2}(x-2)^{2}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dunit0001 View Post
    1) x^2 - x - a - a^2
    Here's another example: factoring by grouping. Rearrange the terms a little.
    x^2 - x - a - a^2 = (x^2 - a^2) + (-x - a)

    Note that the first group factors and the second group factors:
    = (x - a)(x + a) -1 \cdot (x + a)

    Now both the first and the last term have a common factor, x + a, so factor this:
    = ((x - a) - 1)(x + a) = (x - a - 1)(x + a)

    You can use the same trick on 4).

    -Dan
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  4. #4
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    Ok I think I figured it all out now, thanks!
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dunit0001 View Post
    5) 5s^2 - 16s + 12
    You probably should be able to factor this by inspection, or if you can't you can use the ac method. (It's a tutorial, just follow the instructions.)

    I'm going to use a slightly disparaged method of factoring: factoring by solution. I will note that this method is typically only used when the quadratic will not factor over the integers. (Meaning that ax^2 + bx + c does not factor into two linear factors dx + e and fx + g, where a, b, c, d, e, f, and g are all integers.)

    I am showing you this method because I tend to have a mental block when the coefficient of the x^2 term is not 1. When you know the quadratic will factor nicely, you should be able to factor the expression by inspection. As earboth said: Practice!

    Consider the equation
    5s^2 - 16s + 12 = 0

    You don't know how to factor this, and I'm assuming you aren't going to want to go through completing the square (though I recommend doing just that! ), so use the quadratic formula:
    s = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(5)(12)}}{2 \cdot 5}

    etc.

    I get:
    s = 2 and s = \frac{6}{5}.

    Now we need to make factors of this polynomial. Let's start with the s = \frac{6}{5} because it's the harder one.

    First, get rid of the fraction:
    5s = 6

    Now move the constant to the other side of the equation:
    5s - 6 = 0

    What this means is that 5s - 6 is a factor of 5s^2 - 16s + 12 = 0.

    You can find the other factor by doing the same thing with the s = 2 solution.

    -Dan
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