Question

factors : xy^2 -2y^2 +2xy -4y

Answer 1

factor the first two terms and then factor the second two terms, you should have
\[y ^{2}(x-2) +2y(x-2)\]
and look! they both have a (x-2) group in common. Write (x-2)(put what's left outside the parenthesis in here).

Answer 2

What you might notice is that there is the group with the y^2 and 2y have a common factor which you also need to factor out.

Answer 3

(x-2) y (y+2)

Answer 4

Excellent effort. My only criticism(and it's just a small one) is that you put the y out front of the parenthesis instead of in between them and it's all good!

Answer 5

I know what you are saying. The result presented is the output from the Mathematica function, TraditionalForm, a function that cleans up their expression results so they "look right". If the expression x^3 + x^2 + 20 is typed into a Mathematica "notebook", Mathematica spits back:\[20+x^2+x^3 \]If the above is an answer after a lot of this and that, then I normally run the answer through TraditionalForm as shown below:\[\text{TraditionalForm}\left[20+x^2+x^3\right] \to x^3+x^2+20 \]I guess Wolfram in this case, wants to keep the y's together and everything in alphabetical order as shown below:\[\text{TraditionalForm}[y(-2+x)\text{ }(2+y)] \to (x-2) y (y+2) \]\[\text{TraditionalForm}[ y (y+2)(x-2)] \to (x-2) y (y+2) \]

Answer 6

so its ( y^2 -2y) ( x+2) idk cause thats how all the choices are from a through d the answer have to start with Y^2

Answer 7

@jay23,

Where are the choices presented in the problem statement above? "a through d" has no meaning in the context of this problem.

Answer 8

A. (y^2 +2y) (x+2) B (y^2 +2y) (x-2) C ( Y^2-2y) (x+4) D ( y^2 -2y) (x+2)

Answer 9

\[(x-2) y (y+2)=(x-2) \left(2 y+y^2\right)=x y^2+2 x y-2 y^2-4 y \]Choice B:\[(y^2 + 2 y) (x - 2) = x y^2 + 2 x y - 2 y^2 - 4 y \]