Please help (my answer doesn't match book answer)

[kitsae]

8x[sup18ic5mf]2[/sup18ic5mf]y-{3x[sup18ic5mf]2[/sup18ic5mf]y+[2xy[sup18ic5mf]2[/sup18ic5mf]+4x[sup18ic5mf]2[/sup18ic5mf]y-(3xy[sup18ic5mf]2[/sup18ic5mf]-4x[sup18ic5mf]2[/sup18ic5mf]y)]}

The book says the answer is: xy(y-3x)

How?

 

[Loren]

Following the rules to remove grouping symbols, remove the inner-most groupings first, then work outwardly. Next, gather like terms. Finally, factor.

 

[kitsae]

I did that and for some reason my answer is: 5x[sup:d7fr8r52]2[/sup:d7fr8r52]y+xy[sup:d7fr8r52]2[/sup:d7fr8r52]

 

[Loren]

My suggestion is to do it again until you get the correct answer. I did it and the given answer is correct. Figure out what you are doing wrong so you won't do it again. Are you changing the signs inside the grouping symbols when the grouping symbols are preceded by a negative sign?

 

[soroban]

Hello, kitsae!

Obviously, you're making an error . . . Find it!

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

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

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

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

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

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

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

 

[kitsae]

Thank you for your kind assistance. I just have one follow up question. How do you get from the penultimate step to the last step? For example:

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!

 

[mmm4444bot]

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.

 

[Denis]

penultimate

adjective
1. next to the last; "the author inadvertently reveals the murderer in the penultimate chapter"; "the figures in the next-to-last column"

noun
1. the next to last syllable in a word [syn: penult]