Question

1. $(2r^{3}+5r^{2}-2r - 15)div(2r - 3)$

Explanation:

Step1: Use polynomial long division

Divide the first term of the dividend \(2r^{3}\) by the first term of the divisor \(2r\), we get \(r^{2}\). Multiply the divisor \((2r - 3)\) by \(r^{2}\): \(r^{2}(2r - 3)=2r^{3}-3r^{2}\). Subtract this from the dividend:
\[

$$\begin{align*} &(2r^{3}+5r^{2}-2r - 15)-(2r^{3}-3r^{2})\\ =&2r^{3}+5r^{2}-2r - 15 - 2r^{3}+3r^{2}\\ =&8r^{2}-2r - 15 \end{align*}$$

\]

Step2: Divide the new first term

Divide the first term of \(8r^{2}-2r - 15\) (which is \(8r^{2}\)) by the first term of the divisor \(2r\), we get \(4r\). Multiply the divisor \((2r - 3)\) by \(4r\): \(4r(2r - 3)=8r^{2}-12r\). Subtract this from \(8r^{2}-2r - 15\):
\[

$$\begin{align*} &(8r^{2}-2r - 15)-(8r^{2}-12r)\\ =&8r^{2}-2r - 15 - 8r^{2}+12r\\ =&10r - 15 \end{align*}$$

\]

Step3: Divide the new first term again

Divide the first term of \(10r - 15\) (which is \(10r\)) by the first term of the divisor \(2r\), we get \(5\). Multiply the divisor \((2r - 3)\) by \(5\): \(5(2r - 3)=10r - 15\). Subtract this from \(10r - 15\):
\[

$$\begin{align*} &(10r - 15)-(10r - 15)\\ =&10r - 15 - 10r + 15\\ =&0 \end{align*}$$

\]

Step4: Combine the results

The quotient is the sum of the terms we got in each division step: \(r^{2}+4r + 5\), and the remainder is \(0\).

Answer:

\(r^{2}+4r + 5\)