25xto the 3rd degreey to 2degree +125xto 2 degreeyto5degree FACTOR COMPLETELY
\[(25x^3)^2+(125x^2)^5\]?
HOW DO YOU ENTER EXPONENTS ON THE KEYBOARD
i think you have a symol palette where it says \[\sum\]equation
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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