Question 312530
{{{(x^3+3x^2+10x-8)/(3x^2+7x-6)}}}
In order to get {{{x^3}}} multiply {{{3x^2+7x-6}}} by {{{(1/3)x}}} to get {{{x^3+(7/3)x^2-2x}}}.
Subtract that from the original polynomial to get a remainder of,
{{{(x^3+3x^2+10x-8)-(x^3+(7/3)x^2-2x)=(2/3)x^2+12x-8}}}.
Multiply {{{3x^2+7x-6}}} by {{{(2/9)}}} to get {{{(2/3)x^2+(14/9)x-4/3}}}, subtract from the remainder to get the new remainder,
{{{((2/3)x^2+12x-8)-((2/3)x^2+(14/9)x-4/3)= (94/9)x-20/3}}}
No further division can occur,
{{{(x^3+3x^2+10x-8)/(3x^2+7x-6)=(1/3)x+(2/9)+((94/9)x-20/3)/(3x^2+7x-6)}}}