Question

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

Answer 1

\[(25x^3)^2+(125x^2)^5\]?

Answer 2

HOW DO YOU ENTER EXPONENTS ON THE KEYBOARD

Answer 3

i think you have a symol palette where it says \[\sum\]equation

Answer 4

i am writing latex to do it

Answer 5

i would start by writing
\[(5^2x^3)^2+(5^3x^2)^5\]

Answer 6

if this is the problem.

Answer 7

if it is not let me know and i will not continue with this method

Answer 8

NO WHEN I PUT TO WHATEVER DEGREE THAT IS THE EXPONENT

Answer 9

25XTO 3 DEGREE Y TO 2 DEGREE + 125X TO 2 DEGREE Y TO 5 DEGREE

Answer 10

\[25(x^3)^2+125(x^2)^5\]?

Answer 11

YOU LEFT OUT HE Y'S AND THEIR DEGREES
AND NO BRACKETS OR PARENTHESES

Answer 12

oho

Answer 13

\[25x^3y^2+125x^2y^5\]

Answer 14

YES

Answer 15

ok each has a common factor of \[25x^2y^2\] is that clear?

Answer 16

so you can factor it out and write
\[25x^2y^2(x+5y^3\]

Answer 17

NO, I NEED STEP BY STEP

Answer 18

ok.
you have two numbers 25 and 125

Answer 19

YES

Answer 20

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

Answer 21

just looking at that part we can say that
\[25+125=25(1+5)\]

Answer 22

THERE IS NO PARENTHESES

Answer 23

no there are not. but "factoring" means to write as a produce. that is our job, to put the parenthese in

Answer 24

OK

Answer 25

so now for the variables:
first term has \[x^3\] second term has \[x^2\]

Answer 26

their greatest common factor is \[x^2\] because
\[x^2=x^2\times 1\] and \[x^3=x^2\times x\]

Answer 27

I AM SORRY BUT THIS ISN'T HELPING

Answer 28

so we are going to "factor out" a 25 and an \[x^2\]

Answer 29

i have a better idea.

Answer 30

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.

Answer 31

maybe then it will be clear where the
\[25x^2y^2\] came from, and why we pulled it out front of the parentheses