SOLUTION: can some one help me i have done this problem over and over was was told it still needs fixed
a^4b^3 +2a^3b^2 -15a^2=
-15a^2+2ab^2+a^4b^3
Factor out gcf a^2
a^2+(-15 ӏ
Algebra ->
Problems-with-consecutive-odd-even-integers
-> SOLUTION: can some one help me i have done this problem over and over was was told it still needs fixed
a^4b^3 +2a^3b^2 -15a^2=
-15a^2+2ab^2+a^4b^3
Factor out gcf a^2
a^2+(-15 ӏ
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Question 327239: can some one help me i have done this problem over and over was was told it still needs fixed
a^4b^3 +2a^3b^2 -15a^2=
-15a^2+2ab^2+a^4b^3
Factor out gcf a^2
a^2+(-15 〖2a〗^2b+a^(2b^3 ))
a^2 * ((-15 + (2ab^2) + (a^2 * b^3))
this is how i did it and this is what i was told
the GCF you found is not the greatest. You need to consider the b's in common. Then the expression inside parenthesis would have factored further.
can some one help me i keep ending up with the gcf of 2 how do i to the b,s?
You can put this solution on YOUR website! If you copied the problem correctly, "b" is
not part of the gcf.
You are correct in saying the gcf = a^2
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Cheers,
Stan H.
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