Question

(8r² – 31,3 – 14 + 574 -11r) = (r – 6).

Answer 1

ANSWER

$5r^3-r^2\text{ + 2r + 1 remainder -8}$

EXPLANATION

We want to divide:

$8r^2-31r^3-14+5r^4-\text{ 11r}$

by

$r\text{ - 6}$

Let us rearrange:

$5r^4-31r^3+8r^2-\text{ 11r - 14 divided by (r - 6)}$

To do this, we divide each variable (in the long equation) by r(from r - 6), then multiply that by -6 and subtract from the original equation:

$\begin{gathered} (r\text{ - 6) |}5r^4-31r^3+8r^2-\text{ 11r - 14 }\Rightarrow5r^{3\text{ }} \\ \text{ -(5r}^{4\text{ }}-30r^3) \\ \text{ (r - 6) | -r}^3+8r^2-\text{ 11r - 14 }\Rightarrow-r^2 \\ \text{ -(-r}^3+6r^2) \\ (r-6)\text{ |2r}^2\text{ - 11r - 14 }\Rightarrow\text{ 2r} \\ \text{ -(2r}^2\text{ - 12r) } \\ (r-6)\text{ |r - 14 }\Rightarrow\text{ 1} \\ \text{ -(r - 6)} \\ \text{ - 8 (remainder)} \end{gathered}$

Therefore, the answer is:

$5r^3-r^2\text{ + 2r + 1 remainder -8}$