Question

Divide $6a^5+5a^{4}b-8a^3{b}^2-6a^2{b}^3-6a{b}^4$ by $2a^3+3a^{2}b-b^3$ to four terms in the quotient.

Answer 1

Let us solve the given equation by long divisions of polynomial method

$3a^2-2ab-b^2-4a^{-2}{b}^4$

$2a^3+3a^{2}b-b^3)\overline{6a^5+5a^{4}b-8a^3{b}^2-6a^2{b}^3-6a{b}^4}$

$6a^5+9a^{4}b-0a^3{b}^2-3a^2{b}^3$

$\underset{–}{(-)(-)(+)(+)}$

$-4a^{4}b-8a^3{b}^2-3a^2{b}^3-6a{b}^4$

$-4a^{4}b-6a^3{b}^2+0a^2{b}^3+2a{b}^4$

$\underset{–}{(+)(+)(-)(-)}$

$-2a^3{b}^2-3a^2{b}^3-8a{b}^4$

$-2a^3{b}^2-3a^2{b}^3+0a{b}^4+b^5$

$\underset{–}{(+)(+)(-)(-)}$

$-8a{b}^4-b^5$

$-8a{b}^4-12b^5+4a^{-2}{b}^7$

$\underset{–}{(+)(+)(-)}$

$11b^5-4a^{-2}{b}^7$

Since quotient upto four terms is asked, we stop the further division

hence, quotient is $3a^2-2ab-b^2-4a^{-2}{b}^4$

and remainder is $11b^5-4a^{-2}{b}^7$