OpenStudy (anonymous):

How do I solve this? Select one of the factors of x3y2 + 8xy2 - 5x2 - 40 (xy2 + 5) (x2 + 4) (xy2 - 5) (x2 - 8)

3 years ago
OpenStudy (jdoe0001):

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?

3 years ago
OpenStudy (anonymous):

Would it be x and y?

3 years ago
OpenStudy (jdoe0001):

well... what do you think say... for the 1st group is "common" to both terms?

3 years ago
OpenStudy (anonymous):

They both have an x

3 years ago
OpenStudy (jdoe0001):

well yes.... "x" at "some power" though keep in mind that \(\bf x^2\implies xxx\implies x^2x\)

3 years ago
OpenStudy (jdoe0001):

hmm darn typos anyhow \(\bf x^3\implies xxx\implies x^2x\)

3 years ago
OpenStudy (anonymous):

So I change it to that?

3 years ago
OpenStudy (jdoe0001):

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?

3 years ago
OpenStudy (anonymous):

That's where I'm mostly confused, do I write out it out like (xx)(x)(yy)-(5)(xx)?

3 years ago
OpenStudy (jdoe0001):

sure you could do that

3 years ago
OpenStudy (jdoe0001):

\(\bf x^3y^2 + 8xy^2 - 5x^2 - 40\implies (x^3y^2- 5x^2)+(8xy^2- 40) \\ \quad \\ (x^2xy^2- 5x^2)+(8xy^2- 40)\implies (xxxyy-5xx)+(8xyy-40)\) notice any common factors in each group?

3 years ago
OpenStudy (anonymous):

They both have at least one x and two ys?

3 years ago
OpenStudy (jdoe0001):

hmmm well.... the idea is to get a common factor from each group though

3 years ago
OpenStudy (jdoe0001):

say for example.... what common factor could you get from (xxxyy,5xx)?

3 years ago
OpenStudy (anonymous):

Sorry I'm very confused :P

3 years ago