High School

What is the remainder in the synthetic division problem below?

[tex]\[
1 \quad \begin{array}{|ccc}
1 & 2 & -3
\end{array}
\][/tex]

A. 4
B. 3
C. 5
D. 6

Answer :

To solve the synthetic division problem, let's go through the process step-by-step. The synthetic division is used when dividing a polynomial by a binomial of the form [tex]\( x - c \)[/tex].

1. Set Up:
We have the polynomial represented by the coefficients [tex]\([1, 2, -3]\)[/tex] and we are substituting [tex]\( x = 1 \)[/tex].

2. Initialize:
The first number in our division is the leading coefficient, which is [tex]\(1\)[/tex].

3. Synthetic Division Steps:
- Step 1: Bring down the first coefficient [tex]\(1\)[/tex]. This becomes our starting value.
- Step 2: Multiply this first value by [tex]\(1\)[/tex] (our root value) to get [tex]\(1 \times 1 = 1\)[/tex].
- Step 3: Add this result to the next coefficient [tex]\(2\)[/tex]: [tex]\(2 + 1 = 3\)[/tex].
- Step 4: Multiply [tex]\(3\)[/tex] (the new result) by [tex]\(1\)[/tex] to get [tex]\(3 \times 1 = 3\)[/tex].
- Step 5: Add this to the next coefficient [tex]\(-3\)[/tex]: [tex]\(-3 + 3 = 0\)[/tex].

4. Remainder:
The last result, after including all coefficients, is how we determine the remainder. After completing all steps, the remainder in this synthetic division process is [tex]\(0\)[/tex].

Therefore, the remainder of the division problem is [tex]\(0\)[/tex].