bobpursley,math
posted by jas20 .
about this problem , I figured it out and i did get the answer right except that i had forgotten to make it a negative like this:
directions: subtract
4x2x^2+4x^3 from 4x^3+x3x^2

4x^33x^2+x
(4x^32x^2+4x)

so then it becomes
4x^33x^2+x
4x^3+2x^24x

so then 4x^3 and 4x^3 = cancel out
and 3x^2  2x^2 = x^2
and x  (4x)=3x
so the answer is: x^23x
sorry for this part i did a typo:
I said it was this
so then 4x^3 and 4x^3 = cancel out
and 3x^2  2x^2 = x^2
and x  (4x)=3x
BUt it should read like this only for this par:
4x^3 and 4x^3 = cancel out
3+2=1
and
1 + (4)= 3
therefore the answer is: what i said on the previous message:
x^23x
4x2x^2+4x^3 from 4x^3+x3x^2
4x^3+x3x^2
( 4x2x^2+4x^3)
then changing signs and rearranging terms.
4x^3+x3x^2
4x^34x+2x^2
answer after ADDING the above..
3x  x^2
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