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ladyindistress
help (25b^3+15b^2+25b+40)/(5b+5)
I just need a double check please -40b -24/5
I think you should have a b^2 term in there as you have a b^3 in the numerator and only a b term in the denominator.
no it's to the 3rd power
well I'm really confused
it would seem to me you need to factor a 5 out of the top and bottom and then cancel out your 5 so your denomator would be b+5
5b^3+3b^2+5b+8 ---------------- b+1
at that point I would use synthetic division:|dw:1315102006059:dw|
ok so I guess I missed a step
Therefor it should be 5b^2-2b+7+1/(b+5)
(25b^(3)+15b^(2)+25b+40)/(5b+5) Factor out the GCF of 5 from each term in the polynomial. (5(5b^(3))+5(3b^(2))+5(5b)+5(8))/(5b+5) Factor out the GCF of 5 from 25b^(3)+15b^(2)+25b+40. (5(5b^(3)+3b^(2)+5b+8))/(5b+5) Factor out the GCF of 5 from each term in the polynomial. (5(5b^(3)+3b^(2)+5b+8))/(5(b)+5(1)) Factor out the GCF of 5 from 5b+5. (5(5b^(3)+3b^(2)+5b+8))/(5(b+1)) Reduce the expression (5(5b^(3)+3b^(2)+5b+8))/(5(b+1)) by removing a factor of 5 from the numerator and denominator. (5b^(3)+3b^(2)+5b+8)/(b+1)
i am new to open study..... which grade question is this?