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

collect the like terms;

x^{2 }- 4ax - 5a^{2 }= (x+a)(x-5a), not sure how to get there though, probably completing the square

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- February 8th 2013, 05:19 AMndownercollecting like terms; two variables
forgot how to do it, brains have gone to after the summer holiday;

collect the like terms;

x^{2 }- 4ax - 5a^{2 }= (x+a)(x-5a), not sure how to get there though, probably completing the square - February 8th 2013, 05:25 AMjakncokeRe: collecting like terms; two variables
note that if you do x = -a, you get 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 = so b = -5a. - February 8th 2013, 05:34 AMndownerRe: 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

- February 8th 2013, 06:03 AMjakncokeRe: 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 the quadratic formula which gives roots so (x--a)*(x-5a) also gives you the factorization. - February 8th 2013, 11:18 AMearbothRe: collecting like terms; two variables