25xto the 3rd degreey to 2degree +125xto 2 degreeyto5degree FACTOR COMPLETELY

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25xto the 3rd degreey to 2degree +125xto 2 degreeyto5degree FACTOR COMPLETELY

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[(25x^3)^2+(125x^2)^5\]?
HOW DO YOU ENTER EXPONENTS ON THE KEYBOARD
i think you have a symol palette where it says \[\sum\]equation

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Other answers:

i am writing latex to do it
i would start by writing \[(5^2x^3)^2+(5^3x^2)^5\]
if this is the problem.
if it is not let me know and i will not continue with this method
NO WHEN I PUT TO WHATEVER DEGREE THAT IS THE EXPONENT
25XTO 3 DEGREE Y TO 2 DEGREE + 125X TO 2 DEGREE Y TO 5 DEGREE
\[25(x^3)^2+125(x^2)^5\]?
YOU LEFT OUT HE Y'S AND THEIR DEGREES AND NO BRACKETS OR PARENTHESES
oho
\[25x^3y^2+125x^2y^5\]
YES
ok each has a common factor of \[25x^2y^2\] is that clear?
so you can factor it out and write \[25x^2y^2(x+5y^3\]
NO, I NEED STEP BY STEP
ok. you have two numbers 25 and 125
YES
their greatest common factor is 25 because \[25=25\times 1\] and \[125=25\times 5\] so that is going to come out of the parenthese
just looking at that part we can say that \[25+125=25(1+5)\]
THERE IS NO PARENTHESES
no there are not. but "factoring" means to write as a produce. that is our job, to put the parenthese in
OK
so now for the variables: first term has \[x^3\] second term has \[x^2\]
their greatest common factor is \[x^2\] because \[x^2=x^2\times 1\] and \[x^3=x^2\times x\]
I AM SORRY BUT THIS ISN'T HELPING
so we are going to "factor out" a 25 and an \[x^2\]
i have a better idea.
look at my answer which is \[25x^2y^2(x+5y^3)\] multiply out using the distributive law and see if you get what you started with.
maybe then it will be clear where the \[25x^2y^2\] came from, and why we pulled it out front of the parentheses

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