How do I solve this? Select one of the factors of x3y2 + 8xy2 - 5x2 - 40 (xy2 + 5) (x2 + 4) (xy2 - 5) (x2 - 8)
say for example let us group those first \(\bf x^3y^2 + 8xy^2 - 5x^2 - 40\implies (x^3y^2- 5x^2)+(8xy^2- 40)\) now.. can you get a common factor from each group there?
Would it be x and y?
well... what do you think say... for the 1st group is "common" to both terms?
They both have an x
well yes.... "x" at "some power" though keep in mind that \(\bf x^2\implies xxx\implies x^2x\)
hmm darn typos anyhow \(\bf x^3\implies xxx\implies x^2x\)
So I change it to that?
http://www.math-play.com/image-exponents-rules.jpg <--- recall your exponent rules so yes..you can split it like so thus \(\bf x^3y^2 + 8xy^2 - 5x^2 - 40\implies (x^3y^2- 5x^2)+(8xy^2- 40) \\ \quad \\ (x^2xy^2- 5x^2)+(8xy^2- 40)\) so... how would you factor the 1st grup?
That's where I'm mostly confused, do I write out it out like (xx)(x)(yy)-(5)(xx)?
sure you could do that
\(\bf x^3y^2 + 8xy^2 - 5x^2 - 40\implies (x^3y^2- 5x^2)+(8xy^2- 40) \\ \quad \\ (x^2xy^2- 5x^2)+(8xy^2- 40)\implies (xxxyy-5xx)+(8xyy-40)\) notice any common factors in each group?
They both have at least one x and two ys?
hmmm well.... the idea is to get a common factor from each group though
say for example.... what common factor could you get from (xxxyy,5xx)?
Sorry I'm very confused :P
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