Answer :
Sure! Let's perform synthetic division to find the remainder when dividing the polynomial [tex]\(2x^4 - 9x^3 + 23x^2 - 31x + 15\)[/tex] by [tex]\(x-1\)[/tex].
1. Set up the synthetic division:
- Write down the coefficients of the polynomial: [tex]\(2, -9, 23, -31, 15\)[/tex].
- Since we are dividing by [tex]\(x-1\)[/tex], set [tex]\(x - 1 = 0\)[/tex] to find the value of [tex]\(x\)[/tex], which gives us [tex]\(x = 1\)[/tex].
2. Perform the synthetic division process:
- Step 1: Bring down the first coefficient, which is [tex]\(2\)[/tex].
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & & & & \\
\hline
& 2 & & & & \\
\end{array}
\][/tex]
- Step 2: Multiply the value you just brought down (2) by [tex]\(1\)[/tex] (the value of [tex]\(x\)[/tex]), then write the result under the next coefficient.
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & & & \\
\hline
& 2 & & & & \\
\end{array}
\][/tex]
Add this result to the next coefficient [tex]\(-9\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & & & \\
\hline
& 2 & -7 & & & \\
\end{array}
\][/tex]
- Step 3: Repeat this process for each coefficient:
- Multiply [tex]\(-7\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(23\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & & \\
\hline
& 2 & -7 & 16 & & \\
\end{array}
\][/tex]
- Multiply [tex]\(16\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(-31\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & 16 & \\
\hline
& 2 & -7 & 16 & -15 & \\
\end{array}
\][/tex]
- Multiply [tex]\(-15\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(15\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & 16 & -15 \\
\hline
& 2 & -7 & 16 & -15 & 0 \\
\end{array}
\][/tex]
3. Conclusion:
- The final remainder we obtain is [tex]\(0\)[/tex]. This means that [tex]\(x-1\)[/tex] divides the polynomial [tex]\(2x^4 - 9x^3 + 23x^2 - 31x + 15\)[/tex] evenly, with no remainder.
1. Set up the synthetic division:
- Write down the coefficients of the polynomial: [tex]\(2, -9, 23, -31, 15\)[/tex].
- Since we are dividing by [tex]\(x-1\)[/tex], set [tex]\(x - 1 = 0\)[/tex] to find the value of [tex]\(x\)[/tex], which gives us [tex]\(x = 1\)[/tex].
2. Perform the synthetic division process:
- Step 1: Bring down the first coefficient, which is [tex]\(2\)[/tex].
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & & & & \\
\hline
& 2 & & & & \\
\end{array}
\][/tex]
- Step 2: Multiply the value you just brought down (2) by [tex]\(1\)[/tex] (the value of [tex]\(x\)[/tex]), then write the result under the next coefficient.
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & & & \\
\hline
& 2 & & & & \\
\end{array}
\][/tex]
Add this result to the next coefficient [tex]\(-9\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & & & \\
\hline
& 2 & -7 & & & \\
\end{array}
\][/tex]
- Step 3: Repeat this process for each coefficient:
- Multiply [tex]\(-7\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(23\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & & \\
\hline
& 2 & -7 & 16 & & \\
\end{array}
\][/tex]
- Multiply [tex]\(16\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(-31\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & 16 & \\
\hline
& 2 & -7 & 16 & -15 & \\
\end{array}
\][/tex]
- Multiply [tex]\(-15\)[/tex] by [tex]\(1\)[/tex], then add to [tex]\(15\)[/tex]:
[tex]\[
\begin{array}{r|rrrrr}
1 & 2 & -9 & 23 & -31 & 15 \\
& & 2 & -7 & 16 & -15 \\
\hline
& 2 & -7 & 16 & -15 & 0 \\
\end{array}
\][/tex]
3. Conclusion:
- The final remainder we obtain is [tex]\(0\)[/tex]. This means that [tex]\(x-1\)[/tex] divides the polynomial [tex]\(2x^4 - 9x^3 + 23x^2 - 31x + 15\)[/tex] evenly, with no remainder.
