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Answer :
We are given the polynomial
[tex]$$
x^2 + 5x - 14
$$[/tex]
and we want to divide it by [tex]$x - c$[/tex] where in synthetic division the divisor is [tex]$2$[/tex]. This means that [tex]$c = 2$[/tex]. The synthetic division process is as follows:
1. Write down the coefficients of the polynomial: for [tex]$x^2$[/tex], [tex]$x$[/tex], and the constant term, we have [tex]$1$[/tex], [tex]$5$[/tex], and [tex]$-14$[/tex], respectively.
2. Write the value [tex]$2$[/tex] (from the divisor) to the left.
3. Bring down the first coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
& 1 & & \\
\hline
& 1 & &
\end{array}
$$[/tex]
4. Multiply the value just brought down ([tex]$1$[/tex]) by [tex]$2$[/tex]:
[tex]$$
1 \times 2 = 2
$$[/tex]
Place this value under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
& 1 & & \\
\hline
& 1 & &
\end{array}
$$[/tex]
5. Add [tex]$5$[/tex] and [tex]$2$[/tex] to obtain the next coefficient of the quotient term:
[tex]$$
5 + 2 = 7
$$[/tex]
The synthetic division table now looks like:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
& 1 & 7 & \\
\hline
& 1 & 7 &
\end{array}
$$[/tex]
6. Multiply the newly obtained value [tex]$7$[/tex] by [tex]$2$[/tex]:
[tex]$$
7 \times 2 = 14
$$[/tex]
Place it under the next (final) coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & 14 \\
& 1 & 7 & \\
\hline
& 1 & 7 &
\end{array}
$$[/tex]
7. Add [tex]$-14$[/tex] and [tex]$14$[/tex]:
[tex]$$
-14 + 14 = 0
$$[/tex]
8. The result of the synthetic division gives the quotient coefficients and the remainder. The quotient has coefficients [tex]$1$[/tex] and [tex]$7$[/tex], which represents the polynomial
[tex]$$
x + 7,
$$[/tex]
with a remainder of [tex]$0$[/tex].
Thus, the quotient in polynomial form is
[tex]$$
x + 7.
$$[/tex]
The correct answer is: A. [tex]$x+7$[/tex].
[tex]$$
x^2 + 5x - 14
$$[/tex]
and we want to divide it by [tex]$x - c$[/tex] where in synthetic division the divisor is [tex]$2$[/tex]. This means that [tex]$c = 2$[/tex]. The synthetic division process is as follows:
1. Write down the coefficients of the polynomial: for [tex]$x^2$[/tex], [tex]$x$[/tex], and the constant term, we have [tex]$1$[/tex], [tex]$5$[/tex], and [tex]$-14$[/tex], respectively.
2. Write the value [tex]$2$[/tex] (from the divisor) to the left.
3. Bring down the first coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
& 1 & & \\
\hline
& 1 & &
\end{array}
$$[/tex]
4. Multiply the value just brought down ([tex]$1$[/tex]) by [tex]$2$[/tex]:
[tex]$$
1 \times 2 = 2
$$[/tex]
Place this value under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
& 1 & & \\
\hline
& 1 & &
\end{array}
$$[/tex]
5. Add [tex]$5$[/tex] and [tex]$2$[/tex] to obtain the next coefficient of the quotient term:
[tex]$$
5 + 2 = 7
$$[/tex]
The synthetic division table now looks like:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
& 1 & 7 & \\
\hline
& 1 & 7 &
\end{array}
$$[/tex]
6. Multiply the newly obtained value [tex]$7$[/tex] by [tex]$2$[/tex]:
[tex]$$
7 \times 2 = 14
$$[/tex]
Place it under the next (final) coefficient:
[tex]$$
\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & 14 \\
& 1 & 7 & \\
\hline
& 1 & 7 &
\end{array}
$$[/tex]
7. Add [tex]$-14$[/tex] and [tex]$14$[/tex]:
[tex]$$
-14 + 14 = 0
$$[/tex]
8. The result of the synthetic division gives the quotient coefficients and the remainder. The quotient has coefficients [tex]$1$[/tex] and [tex]$7$[/tex], which represents the polynomial
[tex]$$
x + 7,
$$[/tex]
with a remainder of [tex]$0$[/tex].
Thus, the quotient in polynomial form is
[tex]$$
x + 7.
$$[/tex]
The correct answer is: A. [tex]$x+7$[/tex].
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