Question

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

(x2 + 4)

(xy2 - 5)

(x2 - 8)

Answer 1

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?

Answer 2

Would it be x and y?

Answer 3

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

Answer 4

They both have an x

Answer 5

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

Answer 6

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

Answer 7

So I change it to that?

Answer 8

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?

Answer 9

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

Answer 10

sure you could do that

Answer 11

\(\bf x^3y^2 + 8xy^2 - 5x^2 - 40\implies (x^3y^2- 5x^2)+(8xy^2- 40)
\\ \quad \\
(x^2xy^2- 5x^2)+(8xy^2- 40)\implies (xxxyy-5xx)+(8xyy-40)\)

notice any common factors in each group?

Answer 12

They both have at least one x and two ys?

Answer 13

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

Answer 14

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

Answer 15

Sorry I'm very confused :P