- Thread starter kitsae
- Start date

Obviously, you're making an error . . .

\(\displaystyle 8x^2\!y - \bigg\{3x^2\!y+\bigg[2xy^2 + 4x^2\!y - \left(3xy^2 - 4x^2\!y\right) \bigg]\bigg\}\)

\(\displaystyle 8x^2\!y - \bigg\{3x^2\!y + \bigg[2xy^2 + 4x^2\!y - 3xy^2 + 4x^2\!y\bigg]\bigg\}\)

. . \(\displaystyle = \;8x^2\!y - \bigg\{3x^2\!y + 2xy^2 + 4x^2\!y - 3xy^2 + 4x^2\!y\bigg\}\)

. . \(\displaystyle = \;8x^2\!y - \bigg\{11x^2\!y - xy^2\bigg\}\)

. . \(\displaystyle = \;8x^2\!y - 11x^2\!y + xy^2\)

. . \(\displaystyle =\;xy^2 - 3x^2\!y\)

. . \(\displaystyle =\;xy(y-3x)\)

I can get to =xy[sup:1c5fw5sl]2[/sup:1c5fw5sl]-3x[sup:1c5fw5sl]2[/sup:1c5fw5sl]y

How do you get from that to: =xy(y-3x)?

Thank you!

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kitsae said:I can get to =xy[sup:13jt47pu]2[/sup:13jt47pu]-3x[sup:13jt47pu]2[/sup:13jt47pu]y

How do you get from that to: =xy(y-3x)?

Good grief! (Heh, heh.)

That's the easy step!

Both an x and a y are factored out.

If you don't recognize factoring, then think of it as the Distributive Property in reverse.

If that doesn't help, then multiply y - 3x by xy to see that you end up back at the penultimate step.

If you're still at a loss, then go back to your textbook and restudy the sections on factoring.

If that doesn't work, then please let me know or see your instructor.