Answer :
We need to divide the polynomial
[tex]$$2x^2 + 7x + 5$$[/tex]
by the linear divisor
[tex]$$x + 1,$$[/tex]
which means the zero of the divisor is [tex]$-1$[/tex]. The synthetic division process is as follows:
1. Write the coefficients of the polynomial: for [tex]$2x^2$[/tex], [tex]$7x$[/tex], and [tex]$5$[/tex], the coefficients are [tex]$2$[/tex], [tex]$7$[/tex], and [tex]$5$[/tex], respectively.
2. Set up the division using [tex]$-1$[/tex] (since [tex]$x+1=0$[/tex] gives [tex]$x=-1$[/tex]) on the left:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& & &
\end{array}
$$[/tex]
3. Bring down the leading coefficient [tex]$2$[/tex]:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & &
\end{array}
$$[/tex]
4. Multiply the number just written ([tex]$2$[/tex]) by [tex]$-1$[/tex] and write the result under the next coefficient:
[tex]$$2 \times (-1) = -2.$$[/tex]
So we have:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & -2 & \\
\end{array}
$$[/tex]
5. Add the second coefficient [tex]$7$[/tex] and [tex]$-2$[/tex]:
[tex]$$7 + (-2) = 5.$$[/tex]
Write [tex]$5$[/tex] in the row below:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
6. Multiply the number just written ([tex]$5$[/tex]) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1) = -5.$$[/tex]
Place it under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & -5 \\
\end{array}
$$[/tex]
7. Lastly, add the constant term [tex]$5$[/tex] and [tex]$-5$[/tex]:
[tex]$$5 + (-5) = 0.$$[/tex]
So the final row is:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
The numbers in the bottom row (except the remainder) are the coefficients of the quotient. Since we started with a quadratic polynomial, the quotient is one degree lower, namely linear. Therefore, the quotient polynomial is:
[tex]$$2x + 5.$$[/tex]
This result corresponds to option C.
[tex]$$2x^2 + 7x + 5$$[/tex]
by the linear divisor
[tex]$$x + 1,$$[/tex]
which means the zero of the divisor is [tex]$-1$[/tex]. The synthetic division process is as follows:
1. Write the coefficients of the polynomial: for [tex]$2x^2$[/tex], [tex]$7x$[/tex], and [tex]$5$[/tex], the coefficients are [tex]$2$[/tex], [tex]$7$[/tex], and [tex]$5$[/tex], respectively.
2. Set up the division using [tex]$-1$[/tex] (since [tex]$x+1=0$[/tex] gives [tex]$x=-1$[/tex]) on the left:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& & &
\end{array}
$$[/tex]
3. Bring down the leading coefficient [tex]$2$[/tex]:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & &
\end{array}
$$[/tex]
4. Multiply the number just written ([tex]$2$[/tex]) by [tex]$-1$[/tex] and write the result under the next coefficient:
[tex]$$2 \times (-1) = -2.$$[/tex]
So we have:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & -2 & \\
\end{array}
$$[/tex]
5. Add the second coefficient [tex]$7$[/tex] and [tex]$-2$[/tex]:
[tex]$$7 + (-2) = 5.$$[/tex]
Write [tex]$5$[/tex] in the row below:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
6. Multiply the number just written ([tex]$5$[/tex]) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1) = -5.$$[/tex]
Place it under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & -5 \\
\end{array}
$$[/tex]
7. Lastly, add the constant term [tex]$5$[/tex] and [tex]$-5$[/tex]:
[tex]$$5 + (-5) = 0.$$[/tex]
So the final row is:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
The numbers in the bottom row (except the remainder) are the coefficients of the quotient. Since we started with a quadratic polynomial, the quotient is one degree lower, namely linear. Therefore, the quotient polynomial is:
[tex]$$2x + 5.$$[/tex]
This result corresponds to option C.
