Answer :
Sure, let's use synthetic division to test a potential root for the polynomial given by the coefficients [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], [tex]\(c = -12\)[/tex], and [tex]\(d = 0\)[/tex].
The potential root to test is [tex]\( -1 \)[/tex].
The polynomial can be written as:
[tex]\[ P(x) = 1x^3 + 1x^2 - 12x + 0 \][/tex]
Here's a step-by-step solution using synthetic division:
1. Write down the coefficients: List the coefficients of the polynomial in descending order of the power of [tex]\(x\)[/tex]:
[tex]\[
1 \quad 1 \quad -12 \quad 0
\][/tex]
2. Set up the synthetic division: We will use [tex]\( -1 \)[/tex] as the potential root. Write [tex]\( -1 \)[/tex] on the left side and the coefficients on the right side:
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& & & & \\
\end{array}
\][/tex]
3. Bring down the first coefficient: Copy the first coefficient, which is 1, down below the line:
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & & & \\
\end{array}
\][/tex]
4. Multiply and add: Multiply the potential root (-1) by the number just written below the line (1) and write the result under the next coefficient (1):
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & -1 & & \\
\end{array}
\][/tex]
Add this result to the next coefficient (1):
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & & \\
\end{array}
\][/tex]
5. Repeat the multiply and add process:
- For the next position: Multiply [tex]\( -1 \)[/tex] (potential root) by [tex]\( 0 \)[/tex] (last result below the line):
[tex]\[
-1 \times 0 = 0
\][/tex]
Add this to the next coefficient ([tex]\(-12\)[/tex]):
[tex]\[
0 + (-12) = -12
\][/tex]
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & -12 & \\
\end{array}
\][/tex]
- For the final position: Multiply [tex]\( -1 \)[/tex] by [tex]\(-12 \)[/tex]:
[tex]\[
-1 \times -12 = 12
\][/tex]
Add this to the last coefficient (0):
[tex]\[
12 + 0 = 12
\][/tex]
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & -12 & 12 \\
\end{array}
\][/tex]
The numbers that complete the synthetic division problem are:
[tex]\[1, 0, -12, 12\][/tex]
These numbers represent the coefficients of the quotient polynomial and the remainder, showing that we have:
[tex]\[Q(x) = 1x^2 + 0x - 12\][/tex]
and the remainder is [tex]\(12\)[/tex].
Therefore, the final answer after using synthetic division is:
[tex]\[ 1, 0, -12, 12 \][/tex]
The potential root to test is [tex]\( -1 \)[/tex].
The polynomial can be written as:
[tex]\[ P(x) = 1x^3 + 1x^2 - 12x + 0 \][/tex]
Here's a step-by-step solution using synthetic division:
1. Write down the coefficients: List the coefficients of the polynomial in descending order of the power of [tex]\(x\)[/tex]:
[tex]\[
1 \quad 1 \quad -12 \quad 0
\][/tex]
2. Set up the synthetic division: We will use [tex]\( -1 \)[/tex] as the potential root. Write [tex]\( -1 \)[/tex] on the left side and the coefficients on the right side:
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& & & & \\
\end{array}
\][/tex]
3. Bring down the first coefficient: Copy the first coefficient, which is 1, down below the line:
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & & & \\
\end{array}
\][/tex]
4. Multiply and add: Multiply the potential root (-1) by the number just written below the line (1) and write the result under the next coefficient (1):
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & -1 & & \\
\end{array}
\][/tex]
Add this result to the next coefficient (1):
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & & \\
\end{array}
\][/tex]
5. Repeat the multiply and add process:
- For the next position: Multiply [tex]\( -1 \)[/tex] (potential root) by [tex]\( 0 \)[/tex] (last result below the line):
[tex]\[
-1 \times 0 = 0
\][/tex]
Add this to the next coefficient ([tex]\(-12\)[/tex]):
[tex]\[
0 + (-12) = -12
\][/tex]
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & -12 & \\
\end{array}
\][/tex]
- For the final position: Multiply [tex]\( -1 \)[/tex] by [tex]\(-12 \)[/tex]:
[tex]\[
-1 \times -12 = 12
\][/tex]
Add this to the last coefficient (0):
[tex]\[
12 + 0 = 12
\][/tex]
[tex]\[
\begin{array}{r|rrrr}
-1 & 1 & 1 & -12 & 0 \\
\hline
& 1 & 0 & -12 & 12 \\
\end{array}
\][/tex]
The numbers that complete the synthetic division problem are:
[tex]\[1, 0, -12, 12\][/tex]
These numbers represent the coefficients of the quotient polynomial and the remainder, showing that we have:
[tex]\[Q(x) = 1x^2 + 0x - 12\][/tex]
and the remainder is [tex]\(12\)[/tex].
Therefore, the final answer after using synthetic division is:
[tex]\[ 1, 0, -12, 12 \][/tex]
