High School

What is the remainder in the synthetic division problem below?

[tex]\[ 1 \, \bigg| \, 2 \, \, -3 \, \, 2 \][/tex]

A. 3
B. 2
C. 4
D. 5

Answer :

We are asked to find the remainder when dividing the polynomial
[tex]$$
x^3 + 2x^2 - 3x + 2
$$[/tex]
by [tex]$$x - 1$$[/tex] using synthetic division.

Step 1: Write the coefficients of the polynomial:
[tex]$$
1, \; 2, \; -3, \; 2.
$$[/tex]

Step 2: Since the divisor is [tex]$$x - 1$$[/tex], we use [tex]$$1$$[/tex] as the number for synthetic division.

Step 3: Bring the first coefficient down:
[tex]$$
\begin{array}{cccc}
1 & & & \\
\hline
1 & & & \\
\end{array}
$$[/tex]

Step 4: Multiply the number just brought down by [tex]$$1$$[/tex] and write the result under the second coefficient:
[tex]$$
1 \times 1 = 1.
$$[/tex]
Then add this to the second coefficient:
[tex]$$
2 + 1 = 3.
$$[/tex]

Step 5: Repeat the process with the new value:
Multiply [tex]$$3$$[/tex] by [tex]$$1$$[/tex]:
[tex]$$
3 \times 1 = 3.
$$[/tex]
Add this to the third coefficient:
[tex]$$
-3 + 3 = 0.
$$[/tex]

Step 6: Repeat once more with the new value:
Multiply [tex]$$0$$[/tex] by [tex]$$1$$[/tex]:
[tex]$$
0 \times 1 = 0.
$$[/tex]
Add this to the fourth coefficient:
[tex]$$
2 + 0 = 2.
$$[/tex]

The synthetic division process can be summarized in the following table:

[tex]$$
\begin{array}{c|cccc}
\text{Root }1 & 1 & 2 & -3 & 2 \\
\hline
& & 1 & 3 & 0 \\
\hline
& 1 & 3 & 0 & 2 \\
\end{array}
$$[/tex]

The last number, [tex]$$2$$[/tex], is the remainder.

Thus, the remainder when dividing [tex]$$x^3 + 2x^2 - 3x + 2$$[/tex] by [tex]$$x - 1$$[/tex] is [tex]$$\boxed{2}$$[/tex].