Answer :
We are given the polynomial
[tex]$$2x^2 + 7x + 5$$[/tex]
and we need to divide it by
[tex]$$x + 1.$$[/tex]
Notice that since the divisor is in the form [tex]$x+1$[/tex], we take [tex]$r=-1$[/tex] for synthetic division.
Step 1. Write the coefficients of the polynomial:
[tex]$$2, \quad 7, \quad 5.$$[/tex]
Step 2. Set up the synthetic division by writing [tex]$-1$[/tex] to the left:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& & & \\
\end{array}
$$[/tex]
Step 3. Bring down the leading coefficient ([tex]$2$[/tex]):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & & \\
\end{array}
$$[/tex]
Step 4. Multiply this number ([tex]$2$[/tex]) by [tex]$-1$[/tex]:
[tex]$$2 \times (-1)= -2.$$[/tex]
Write this number under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & \\
\hline
& 2 & & \\
\end{array}
$$[/tex]
Step 5. Add vertically in the second column:
[tex]$$7 + (-2) = 5.$$[/tex]
So we write [tex]$5$[/tex] in the bottom row:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 6. Multiply [tex]$5$[/tex] (the new value) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1)= -5.$$[/tex]
Write this under the third coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & -5 \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 7. Add vertically in the third column:
[tex]$$5 + (-5) = 0.$$[/tex]
This is the remainder:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & -5 \\
\hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
Since the remainder is [tex]$0$[/tex], we have an exact division. The numbers in the bottom row (besides the remainder) give the coefficients of the quotient polynomial.
Since the original polynomial was quadratic ([tex]$2x^2+7x+5$[/tex]), the quotient will be linear:
- The first number corresponds to [tex]$2x$[/tex], and
- The second number is [tex]$+5$[/tex].
Thus, the quotient is:
[tex]$$2x + 5.$$[/tex]
The correct choice is [tex]$\boxed{2x+5}$[/tex].
[tex]$$2x^2 + 7x + 5$$[/tex]
and we need to divide it by
[tex]$$x + 1.$$[/tex]
Notice that since the divisor is in the form [tex]$x+1$[/tex], we take [tex]$r=-1$[/tex] for synthetic division.
Step 1. Write the coefficients of the polynomial:
[tex]$$2, \quad 7, \quad 5.$$[/tex]
Step 2. Set up the synthetic division by writing [tex]$-1$[/tex] to the left:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& & & \\
\end{array}
$$[/tex]
Step 3. Bring down the leading coefficient ([tex]$2$[/tex]):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
\hline
& 2 & & \\
\end{array}
$$[/tex]
Step 4. Multiply this number ([tex]$2$[/tex]) by [tex]$-1$[/tex]:
[tex]$$2 \times (-1)= -2.$$[/tex]
Write this number under the next coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & \\
\hline
& 2 & & \\
\end{array}
$$[/tex]
Step 5. Add vertically in the second column:
[tex]$$7 + (-2) = 5.$$[/tex]
So we write [tex]$5$[/tex] in the bottom row:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 6. Multiply [tex]$5$[/tex] (the new value) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1)= -5.$$[/tex]
Write this under the third coefficient:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & -5 \\
\hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 7. Add vertically in the third column:
[tex]$$5 + (-5) = 0.$$[/tex]
This is the remainder:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\
& & -2 & -5 \\
\hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
Since the remainder is [tex]$0$[/tex], we have an exact division. The numbers in the bottom row (besides the remainder) give the coefficients of the quotient polynomial.
Since the original polynomial was quadratic ([tex]$2x^2+7x+5$[/tex]), the quotient will be linear:
- The first number corresponds to [tex]$2x$[/tex], and
- The second number is [tex]$+5$[/tex].
Thus, the quotient is:
[tex]$$2x + 5.$$[/tex]
The correct choice is [tex]$\boxed{2x+5}$[/tex].
