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Mathematics 14 Online
OpenStudy (anonymous):

Please help with the following answers. Factor 33x^2y^4z + 55xy^2z^5 + 66x^yz^3

OpenStudy (anonymous):

It's not meant to me x^y, right? Otherwise factorising that would be... fun (read: impossible, apart from the factor of 11)

OpenStudy (anonymous):

be*

OpenStudy (anonymous):

33x^2y^4z+55xy^2z^5+66^6yz^3

OpenStudy (anonymous):

Assume it's 66x^6yz^3 at the end, then you need to pull out the common factor of 11, and xyz (the highest power of x, y, and z in all is 1) Answer is of the form 11xyz( blah)

OpenStudy (anonymous):

what is the anwser

OpenStudy (anonymous):

If you want the answer, go check it on any online calculator worth its bandwidth. If you want advice and help to get the answer yourself, then ask here.

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