Question

1. $(y^4 - y^2 + 8y + 5) div (y + 1)$

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(y^4 - y^2 + 8y + 5\) by the leading term of the divisor \(y + 1\). The leading term of the dividend is \(y^4\) and the leading term of the divisor is \(y\), so \(y^4\div y = y^3\). Multiply the divisor \(y + 1\) by \(y^3\) to get \(y^4 + y^3\). Subtract this from the dividend:
\[

$$\begin{align*} &(y^4 - y^2 + 8y + 5)-(y^4 + y^3)\\ =&y^4 - y^2 + 8y + 5 - y^4 - y^3\\ =& -y^3 - y^2 + 8y + 5 \end{align*}$$

\]

Step2: Divide the new leading term

The new leading term is \(-y^3\), divide by \(y\) to get \(-y^2\). Multiply the divisor by \(-y^2\) to get \(-y^3 - y^2\). Subtract this from the previous result:
\[

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

\]

Step3: Divide the new leading term

The new leading term is \(8y\), divide by \(y\) to get \(8\). Multiply the divisor by \(8\) to get \(8y + 8\). Subtract this from the previous result:
\[

$$\begin{align*} &(8y + 5)-(8y + 8)\\ =& 8y + 5 - 8y - 8\\ =& -3 \end{align*}$$

\]

Step4: Combine the results

The quotient is \(y^3 - y^2 + 8\) and the remainder is \(-3\). So we can write the division as:
\[
\frac{y^4 - y^2 + 8y + 5}{y + 1}=y^3 - y^2 + 8-\frac{3}{y + 1}
\]
Or, in quotient - remainder form: \((y^4 - y^2 + 8y + 5)=(y + 1)(y^3 - y^2 + 8)-3\)

Answer:

The quotient is \(y^3 - y^2 + 8\) and the remainder is \(-3\), so \((y^4 - y^2 + 8y + 5)\div(y + 1)=y^3 - y^2 + 8-\frac{3}{y + 1}\) (or in quotient - remainder form \((y + 1)(y^3 - y^2 + 8)-3\))