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Math Help - collecting like terms; two variables

  1. #1
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    collecting like terms; two variables

    forgot how to do it, brains have gone to after the summer holiday;

    collect the like terms;

    x2 - 4ax - 5a2

    = (x+a)(x-5a), not sure how to get there though, probably completing the square
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: collecting like terms; two variables

    note that if you do x = -a, you get  a^2 + 4a^2 - 5a^2 = 0 so you can factor out (x--a) or (x+a).

    so (x+a)*(x-5a) , because if you can factor out (x+a), and you do (x+a)*(x+b), then a*b =  -5a^2 so b = -5a.
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  3. #3
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    Re: collecting like terms; two variables

    i appreciate the quick response, but I have no idea how you factor out (x+a) from a^2 + 4a^2 - 5a^2
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: collecting like terms; two variables

    For any polynomial p(x), if you can find a root r, so p(r) = 0, then (x-r) divides p(x) with remainder 0. so you get p(x) = (x-r)*quotient .

    but in your case using the quadratic formula can also solve this dilemna.

    So for  x^2 - 4ax - 5a^2 the quadratic formula  \frac{4a \pm \sqrt{16a^2 - 4(1)(-5a^2)}}{2} = \frac{4a \pm \sqrt{36a^2}}{2} which gives roots  x_1 = -a, x_2 = 5a so (x--a)*(x-5a) also gives you the factorization.
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  5. #5
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    Re: collecting like terms; two variables

    Quote Originally Posted by ndowner View Post
    forgot how to do it, brains have gone to after the summer holiday;

    collect the like terms;

    x2 - 4ax - 5a2

    = (x+a)(x-5a), not sure how to get there though, probably completing the square
    This is a slightly different way:

    x^2-4ax-5a^2 = x^2-4ax\color{red}+\left(\frac{4a}2 \right)^2 - 4a^2\color{black}-5a^2 = \underbrace{(x-2a)^2-9a^2}_{\text{difference of squares}}

    You can factor a difference of squares to:

    (x-2a)^2-9a^2 = \left( (x-2a)-3a\right)\left( (x-2a)+3a\right) = \boxed{(x-5a)(x+a)}
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