Question

use synthetic division to find the result when $3x^4 - 17x^3 - 24x^2 - 26x - 14$ is divided by $x - 7$.

Explanation:

Step1: Set up synthetic division

For dividing by \(x - 7\), we use \(7\) as the root. The coefficients of the polynomial \(3x^{4}-17x^{3}-24x^{2}-26x - 14\) are \(3\), \(-17\), \(-24\), \(-26\), \(-14\).
Set up the synthetic division as:
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & & & & \\ \hline & 3 & & & & \\ \end{array}$$

\]

Step2: Bring down the leading coefficient

Bring down the \(3\) to the bottom row:
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & & & & \\ \hline & 3 & & & & \\ \end{array}$$

\]

Step3: Multiply and add

Multiply \(7\) by \(3 = 21\), add to \(-17\): \(-17+21 = 4\)
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & 21 & & & \\ \hline & 3 & 4 & & & \\ \end{array}$$

\]
Multiply \(7\) by \(4=28\), add to \(-24\): \(-24 + 28=4\)
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & 21 & 28 & & \\ \hline & 3 & 4 & 4 & & \\ \end{array}$$

\]
Multiply \(7\) by \(4 = 28\), add to \(-26\): \(-26+28 = 2\)
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & 21 & 28 & 28 & \\ \hline & 3 & 4 & 4 & 2 & \\ \end{array}$$

\]
Multiply \(7\) by \(2=14\), add to \(-14\): \(-14 + 14=0\)
\[

$$\begin{array}{r|rrrrr} 7 & 3 & -17 & -24 & -26 & -14 \\ & & 21 & 28 & 28 & 14 \\ \hline & 3 & 4 & 4 & 2 & 0 \\ \end{array}$$

\]
The bottom row coefficients (excluding the last \(0\) which is the remainder) are \(3\), \(4\), \(4\), \(2\). So the quotient polynomial is \(3x^{3}+4x^{2}+4x + 2\) with a remainder of \(0\).

Answer:

The result of the division is \(3x^{3}+4x^{2}+4x + 2\) (since the remainder is \(0\), we can say \(3x^{4}-17x^{3}-24x^{2}-26x - 14=(x - 7)(3x^{3}+4x^{2}+4x + 2)\))