## ladyindistress 4 years ago help (25b^3+15b^2+25b+40)/(5b+5)

I just need a double check please -40b -24/5

2. TBates

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

5. zbay

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

6. TBates

*b+1

7. zbay

8. zbay

5b^3+3b^2+5b+8 ---------------- b+1

9. TBates

at that point I would use synthetic division:|dw:1315102006059:dw|

ok so I guess I missed a step

11. TBates

Therefor it should be 5b^2-2b+7+1/(b+5)

12. angel98

5b^3+3b^2+5b+8/b+1

13. angel98

(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)

14. mujtabazarrar

i am new to open study..... which grade question is this?