Question

use synthetic division to find the remainder.\\(\frac{x^4 - x^3 + 5x^2 - 127x - 7}{x - 5}\\)

Explanation:

Step1: Identify the root and coefficients

For synthetic division with divisor \(x - 5\), the root is \(r = 5\). The coefficients of the dividend \(x^{4}-x^{3}+5x^{2}-127x - 7\) are \(1, - 1,5,-127,-7\) (for \(x^{4},x^{3},x^{2},x,\) and constant term respectively).

Step2: Set up synthetic division

Write down the root \(5\) and the coefficients:
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & & & & \\ \hline & & & & & \\ \end{array}$$

\]

Step3: Bring down the first coefficient

Bring down the first coefficient \(1\):
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & & & & \\ \hline & 1 & & & & \\ \end{array}$$

\]

Step4: Multiply and add

Multiply the brought - down number (\(1\)) by the root (\(5\)): \(1\times5 = 5\). Add this to the next coefficient (\(-1\)): \(-1 + 5=4\).
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & 5 & & & \\ \hline & 1 & 4 & & & \\ \end{array}$$

\]
Multiply the new number (\(4\)) by the root (\(5\)): \(4\times5 = 20\). Add this to the next coefficient (\(5\)): \(5+20 = 25\).
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & 5 & 20 & & \\ \hline & 1 & 4 & 25 & & \\ \end{array}$$

\]
Multiply the new number (\(25\)) by the root (\(5\)): \(25\times5=125\). Add this to the next coefficient (\(-127\)): \(-127 + 125=-2\).
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & 5 & 20 & 125 & \\ \hline & 1 & 4 & 25 & -2 & \\ \end{array}$$

\]
Multiply the new number (\(-2\)) by the root (\(5\)): \(-2\times5=-10\). Add this to the last coefficient (\(-7\)): \(-7+( - 10)=-17\).
\[

$$\begin{array}{r|rrrrr} 5 & 1 & -1 & 5 & -127 & -7 \\ & & 5 & 20 & 125 & -10 \\ \hline & 1 & 4 & 25 & -2 & -17 \\ \end{array}$$

\]

The last number in the bottom row is the remainder.

Answer:

The remainder is \(-17\)