Complete the synthetic division problem below.

\[ 2 \left\lvert\, \begin{array}{lll}
\hline 1 & 5 & -14
\end{array}\right. \]

What is the quotient in polynomial form?

A. \( x+7 \)

B. \( x-7 \)

C. \( x-5 \)

D. \( x+5 \)

Answer :

Sure! Let's work through this synthetic division problem step-by-step.

We are given the polynomial [tex]\( x^2 + 5x - 14 \)[/tex] and are using synthetic division to divide it by [tex]\( x - 2 \)[/tex].

The coefficients of the polynomial are: [tex]\( 1 \)[/tex] (for [tex]\( x^2 \)[/tex]), [tex]\( 5 \)[/tex] (for [tex]\( x \)[/tex]), and [tex]\( -14 \)[/tex] (the constant term).

The divisor here is [tex]\( x - 2 \)[/tex], which means we'll be using the value [tex]\( 2 \)[/tex] in our synthetic division.

1. Setup: Write down the coefficients of the polynomial:
```
1 5 -14
```
Next, we use the value [tex]\( 2 \)[/tex] (since we're dividing by [tex]\( x - 2 \)[/tex]).

2. Process: Start with the leading coefficient:
- Bring down the [tex]\( 1 \)[/tex] as is.

3. Multiply and Add:
- Multiply the [tex]\( 1 \)[/tex] by [tex]\( 2 \)[/tex] (our synthetic divisor), which gives [tex]\( 2 \)[/tex].
- Add this [tex]\( 2 \)[/tex] to the next coefficient (which is [tex]\( 5 \)[/tex]):
[tex]\[
5 + 2 = 7
\][/tex]

4. Repeat:
- Multiply the [tex]\( 7 \)[/tex] by [tex]\( 2 \)[/tex] (our synthetic divisor), which gives [tex]\( 14 \)[/tex].
- Add this [tex]\( 14 \)[/tex] to the last coefficient (which is [tex]\(-14\)[/tex]):
[tex]\[
-14 + 14 = 0
\][/tex]

5. Result: The numbers left at the bottom are the coefficients of the quotient. In this case, we're left with:
```
1 7
```
The last number, [tex]\( 0 \)[/tex], is the remainder.

Thus, the quotient polynomial is [tex]\( x + 7 \)[/tex].

So, the resulting polynomial is:
[tex]\[ x + 7 \][/tex]

Therefore, the correct answer is A. [tex]\( x + 7 \)[/tex].