Answer :
We wish to divide the polynomial
[tex]$$
x^2 + 5x - 14
$$[/tex]
by
[tex]$$
x - 2.
$$[/tex]
Using synthetic division with the divisor value [tex]$2$[/tex], we follow these steps:
1. Write the coefficients of the polynomial: for [tex]$x^2$[/tex], [tex]$x$[/tex], and the constant term, the coefficients are [tex]$1$[/tex], [tex]$5$[/tex], and [tex]$-14$[/tex], respectively.
2. Set up the synthetic division by writing [tex]$2$[/tex] on the left and the coefficients in a row:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
\hline
& & &
\end{array}$$[/tex]
3. Bring down the first coefficient:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
\hline
& 1 & &
\end{array}$$[/tex]
4. Multiply the number just brought down ([tex]$1$[/tex]) by [tex]$2$[/tex]:
[tex]$$
1 \times 2 = 2.
$$[/tex]
5. Write this result under the second coefficient and add:
[tex]$$5 + 2 = 7.$$[/tex]
The table now looks like:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
\hline
& 1 & 7 &
\end{array}$$[/tex]
6. Multiply the new result ([tex]$7$[/tex]) by [tex]$2$[/tex]:
[tex]$$
7 \times 2 = 14.
$$[/tex]
7. Write this number under the third coefficient and add:
[tex]$$-14 + 14 = 0.$$[/tex]
The completed table is:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & 14 \\
\hline
& 1 & 7 & 0
\end{array}$$[/tex]
8. The bottom row now represents the coefficients of the quotient polynomial and the remainder. Here, [tex]$1$[/tex] and [tex]$7$[/tex] are the coefficients for [tex]$x$[/tex] and the constant term, respectively, and the final [tex]$0$[/tex] is the remainder.
Thus, the quotient polynomial is
[tex]$$
x + 7.
$$[/tex]
Since the quotient in polynomial form is [tex]$x+7$[/tex], the correct answer is:
D. [tex]$x+7$[/tex].
[tex]$$
x^2 + 5x - 14
$$[/tex]
by
[tex]$$
x - 2.
$$[/tex]
Using synthetic division with the divisor value [tex]$2$[/tex], we follow these steps:
1. Write the coefficients of the polynomial: for [tex]$x^2$[/tex], [tex]$x$[/tex], and the constant term, the coefficients are [tex]$1$[/tex], [tex]$5$[/tex], and [tex]$-14$[/tex], respectively.
2. Set up the synthetic division by writing [tex]$2$[/tex] on the left and the coefficients in a row:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
\hline
& & &
\end{array}$$[/tex]
3. Bring down the first coefficient:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & & \\
\hline
& 1 & &
\end{array}$$[/tex]
4. Multiply the number just brought down ([tex]$1$[/tex]) by [tex]$2$[/tex]:
[tex]$$
1 \times 2 = 2.
$$[/tex]
5. Write this result under the second coefficient and add:
[tex]$$5 + 2 = 7.$$[/tex]
The table now looks like:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & \\
\hline
& 1 & 7 &
\end{array}$$[/tex]
6. Multiply the new result ([tex]$7$[/tex]) by [tex]$2$[/tex]:
[tex]$$
7 \times 2 = 14.
$$[/tex]
7. Write this number under the third coefficient and add:
[tex]$$-14 + 14 = 0.$$[/tex]
The completed table is:
[tex]$$\begin{array}{r|ccc}
2 & 1 & 5 & -14 \\
& & 2 & 14 \\
\hline
& 1 & 7 & 0
\end{array}$$[/tex]
8. The bottom row now represents the coefficients of the quotient polynomial and the remainder. Here, [tex]$1$[/tex] and [tex]$7$[/tex] are the coefficients for [tex]$x$[/tex] and the constant term, respectively, and the final [tex]$0$[/tex] is the remainder.
Thus, the quotient polynomial is
[tex]$$
x + 7.
$$[/tex]
Since the quotient in polynomial form is [tex]$x+7$[/tex], the correct answer is:
D. [tex]$x+7$[/tex].
