Question

9x^2+9x^3-7-13x/3x+4

Answer 1

3x+4)9x^3+9x^2-13x-7(3x^2-x-1/3
9x^3+12x^2
-----------------------
-3x^2-13x-7
-3x^2-12x
---------------------
-x-7
-x-4/3
-------------------------
-17/3

Answer 2

here 3x^2-x-1/3 is the quotient and -17/3 is remainder

Answer 3

\[(3x ^{2}-3x-1/3) -17/(9x+12) \]

Answer 4

\[\frac{9x^2+9x^3-7-13x}{3x+4}=\frac{-9x^3+10x-7}{4}\]

Answer 5

how Craigwillium?

Answer 6

can u please show the steps...

Answer 7

It's a matter of simplification by moving around exponents. I only did the simplification part.

Answer 8

@craig Oh even i em interested in seeing the proof