Answer :
We are given the synthetic division setup for dividing the polynomial
[tex]$$P(x) = 1x^3 + 2x^2 - 3x + 2$$[/tex]
by the divisor corresponding to [tex]$x - 1$[/tex], which has a zero at [tex]$x = 1$[/tex].
Here is a step-by-step explanation:
1. Write down the coefficients of the polynomial:
[tex]$$1,\quad 2,\quad -3,\quad 2.$$[/tex]
2. Place the zero of the divisor, [tex]$1$[/tex], to the left:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & & & \\
\hline
& & & &
\end{array}$$[/tex]
3. Bring down the first coefficient, [tex]$1$[/tex], to the bottom row:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & & & \\
\hline
& 1 & & &
\end{array}$$[/tex]
4. Multiply the number in the bottom row by [tex]$1$[/tex] (the zero of the divisor) and write the result underneath the next coefficient:
[tex]$$1 \times 1 = 1.$$[/tex]
Then add this to the next coefficient, [tex]$2$[/tex], to get:
[tex]$$2 + 1 = 3.$$[/tex]
The table now looks like:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & & \\
\hline
& 1 & 3 & &
\end{array}$$[/tex]
5. Multiply the new bottom row term by [tex]$1$[/tex]:
[tex]$$3 \times 1 = 3.$$[/tex]
Add this to the third coefficient, [tex]$-3$[/tex], to obtain:
[tex]$$-3 + 3 = 0.$$[/tex]
Updated table:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & 3 & \\
\hline
& 1 & 3 & 0 &
\end{array}$$[/tex]
6. Multiply the latest term by [tex]$1$[/tex]:
[tex]$$0 \times 1 = 0.$$[/tex]
Finally, add this to the last coefficient, [tex]$2$[/tex], giving:
[tex]$$2 + 0 = 2.$$[/tex]
The completed table is:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & 3 & 0 \\
\hline
& 1 & 3 & 0 & 2
\end{array}$$[/tex]
The last number on the bottom row is the remainder. Therefore, the remainder is
[tex]$$\boxed{2}.$$[/tex]
[tex]$$P(x) = 1x^3 + 2x^2 - 3x + 2$$[/tex]
by the divisor corresponding to [tex]$x - 1$[/tex], which has a zero at [tex]$x = 1$[/tex].
Here is a step-by-step explanation:
1. Write down the coefficients of the polynomial:
[tex]$$1,\quad 2,\quad -3,\quad 2.$$[/tex]
2. Place the zero of the divisor, [tex]$1$[/tex], to the left:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & & & \\
\hline
& & & &
\end{array}$$[/tex]
3. Bring down the first coefficient, [tex]$1$[/tex], to the bottom row:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & & & \\
\hline
& 1 & & &
\end{array}$$[/tex]
4. Multiply the number in the bottom row by [tex]$1$[/tex] (the zero of the divisor) and write the result underneath the next coefficient:
[tex]$$1 \times 1 = 1.$$[/tex]
Then add this to the next coefficient, [tex]$2$[/tex], to get:
[tex]$$2 + 1 = 3.$$[/tex]
The table now looks like:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & & \\
\hline
& 1 & 3 & &
\end{array}$$[/tex]
5. Multiply the new bottom row term by [tex]$1$[/tex]:
[tex]$$3 \times 1 = 3.$$[/tex]
Add this to the third coefficient, [tex]$-3$[/tex], to obtain:
[tex]$$-3 + 3 = 0.$$[/tex]
Updated table:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & 3 & \\
\hline
& 1 & 3 & 0 &
\end{array}$$[/tex]
6. Multiply the latest term by [tex]$1$[/tex]:
[tex]$$0 \times 1 = 0.$$[/tex]
Finally, add this to the last coefficient, [tex]$2$[/tex], giving:
[tex]$$2 + 0 = 2.$$[/tex]
The completed table is:
[tex]$$\begin{array}{r|rrrr}
1 & 1 & 2 & -3 & 2 \\
& & 1 & 3 & 0 \\
\hline
& 1 & 3 & 0 & 2
\end{array}$$[/tex]
The last number on the bottom row is the remainder. Therefore, the remainder is
[tex]$$\boxed{2}.$$[/tex]
