Answer
a) The possible rational zeroes for the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ are $\pm 1,\pm 2,\pm 4$.
b)
An actual zero for the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ is $2$.
c) The remaining zeroes of the polynomial function are $-1\text{ and }-2$.
Work Step by Step
(a)
Here, the constant term is $-4$ and the leading coefficient is 1.
The factors of the constant term, $-4$ are $\pm 1,\pm 2,\pm 4$ and the factors of the leading coefficient,1 are $\pm 1$.
So, the list of all possible rational zeroes is calculated by the formula:
$\begin{align}
& \text{Possible rational zeroes}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \\
& =\frac{\text{Factors of }-4}{\text{Factors of 1}} \\
& =\frac{\pm 1,\pm 2,\pm 4}{\pm 1} \\
& =\pm 1,\pm 2,\pm 4
\end{align}$
Therefore, there are total six possible rational zeroes for the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ that are $\pm 1,\pm 2,\pm 4$.
(b)
If the function $f\left( x \right)$ is divided by $x-c$ and the remainder is zero then this means that c is a zero of the function.
Now, use synthetic division to find a rational zero among the list of possible rational zeroes $\pm 1,\pm 2,\pm 4$.
Start by testing if 1 is a rational zero. If it is not a rational zero, then test other possible rational zeroes.
Now write 1 for the divisor and write the coefficients of the dividend as shown below,
$\left. {\underline {\,
1 \,}}\! \right| \text{ 1 1 }-4\text{ }-4$
Now bring down the leading coefficient of the dividend on the bottom row as shown below,
$\begin{align}
& \left. {\underline {\,
1 \,}}\! \right| \text{ 1 1 }-4\text{ }-4 \\
& \text{ }\underline{\downarrow \text{Bring down 1}} \\
& \text{ 1} \\
\end{align}$
Multiply 1 with the bottom row value and write the product in the next column in the second row as shown below,
$\begin{align}
& \left. {\underline {\,
1 \,}}\! \right| \text{ 1 1 }-4\text{ }-4 \\
& \text{ }\underline{\downarrow \text{ 1}} \\
& \text{ 1} \\
\end{align}$
Add the column values and write down the sum in the bottom row as shown below,
$\begin{align}
& \left. {\underline {\,
1 \,}}\! \right| \text{ 1 1 }-4\text{ }-4 \\
& \text{ }\underline{\downarrow \text{ 1}} \\
& \text{ 1 2} \\
\end{align}$
Similarly repeat this series of multiplications and additions until the columns are filled in,
$\begin{align}
& \left. {\underline {\,
1 \,}}\! \right| \text{ 1 1 }-4\text{ }-4 \\
& \text{ }\underline{\downarrow \text{ 1 2 }-2} \\
& \text{ 1 2 }-\text{2 }-6 \\
\end{align}$
Here, the reminder is $-6$. This means 1 is not a zero of the function.
Now, test for 2 if it is a rational zero.
$\begin{align}
& \left. {\underline {\,
2 \,}}\! \right| \text{ 1 1 }-4\text{ }-4 \\
& \text{ }\underline{\downarrow \text{ 2 6 4}} \\
& \text{ 1 3 2 0} \\
\end{align}$
Here, the reminder is $0$. This means 2 is a zero of the function and the first three numbers $1,3,2$ in the bottom row give the coefficients of the quotient.
Now, the quotient is ${{x}^{2}}+3x+2$.
Hence, the actual zero for the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ is $2$.
(c)
If the function $f\left( x \right)$ is divided by $x-c$ and the remainder is zero then this means c is a zero of the function.
So, $x-2$ is a factor of the polynomial and from part (b), the quotient is ${{x}^{2}}+3x+2$.
Now, the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ can be written as,
$f\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+3x+2 \right)$
Now, factorize the quotient by the middle term splitting method.
That is,
$\begin{align}
& f\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+3x+2 \right) \\
& =\left( x-2 \right)\left( {{x}^{2}}+2x+x+2 \right) \\
& =\left( x-2 \right)\left\{ x\left( x+2 \right)+1\left( x+2 \right) \right\} \\
& =\left( x-2 \right)\left( x+2 \right)\left( x+1 \right)
\end{align}$
Set each factor equal to zero to find the remaining zeroes of the function.
That is,
$\begin{align}
& f\left( x \right)=0 \\
& \left( x-2 \right)\left( x+2 \right)\left( x+1 \right)=0
\end{align}$
Which gives $x=-1,2,-2$.
Hence, the remaining zeroes of the polynomial function are $-1\text{ and }-2$.
