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Mathematics 14 Online
OpenStudy (anonymous):

Select one of the factors of x2y3 - 11x2y + 6y2 - 66. (x2y + 6) (x2y - 6) (y2 + 11) (y - 3) HELP!!!!

hartnn (hartnn):

consider \(\Large x^2 y^3 +6y^2\) what can you factor out from these 2 terms ?

OpenStudy (anonymous):

okay hold on

OpenStudy (anonymous):

\[x ^{2}y ^{3} + 6y ^{2} \]

OpenStudy (anonymous):

hold on im still going

hartnn (hartnn):

holding on :)

OpenStudy (anonymous):

\[y ^{2} \left( x ^{2} +6y \right)\]

OpenStudy (anonymous):

now am i suppose to factor 6y?

hartnn (hartnn):

you correctly figured out that y^2 is common and can be factored but you didn't factor out y^2 correctly

OpenStudy (anonymous):

i dont understand

hartnn (hartnn):

\(\Large x^2 y^3 = y^2 \times (x^2 y)\) ^^ got that ?

hartnn (hartnn):

\(\Large 6y^2 = y^2 \times (6)\)

OpenStudy (anonymous):

but what about the 6?

OpenStudy (anonymous):

oh neva mind

hartnn (hartnn):

tell me if you have any doubts, make sure you understand whats going on ...

hartnn (hartnn):

\(\Large x^2y^3 +6y^2 = y^2 (x^2 y)+ y^2 (6) \\ \Large \color{red} = y^2 (x^2y+6)\)

hartnn (hartnn):

makes sense ?

OpenStudy (anonymous):

oh yeas thank you i know what i did wrong i put the y with the six when i wasnt suppose to because it was already factored out correct?

hartnn (hartnn):

correct :) and you also missed one 'y' with x^2, because only 2 y's out of y^3 was factored, one still remains

OpenStudy (anonymous):

oh okay so the correct answer would be a?

hartnn (hartnn):

thats right :) A is correct option.

OpenStudy (anonymous):

thank you so much i have other questions but i want to make sure that you get medals and stuff so ill keep posting will you be online?

hartnn (hartnn):

i'll help until i am online :) and welcome ^_^

OpenStudy (anonymous):

okay :)

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