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 when performing synthetic division, we use the root of the divisor. Since
[tex]$$x + 1 = 0 \quad \Longrightarrow \quad x = -1,$$[/tex]
we will use [tex]$-1$[/tex] in our division steps.
Step 1. Write the coefficients of the dividend polynomial in order:
[tex]$$2, \quad 7, \quad 5.$$[/tex]
Step 2. Set up the synthetic division by writing [tex]$-1$[/tex] on the left and the coefficients to its right:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& & & \\
\end{array}
$$[/tex]
Step 3. Bring down the first coefficient (2):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & & \\
\end{array}
$$[/tex]
Step 4. Multiply the number you just brought down (2) by [tex]$-1$[/tex]:
[tex]$$2 \times (-1) = -2.$$[/tex]
Write this number under the next coefficient (7):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & -2 & \\
\end{array}
$$[/tex]
Then, add [tex]$7$[/tex] and [tex]$-2$[/tex]:
[tex]$$7 + (-2) = 5.$$[/tex]
Write the result (5) below the line:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 5. Multiply the new value (5) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1) = -5.$$[/tex]
Place this number under the last coefficient (5):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & -5 \\
\end{array}
$$[/tex]
Now, add the last column:
[tex]$$5 + (-5) = 0.$$[/tex]
Write the 0 as the remainder:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
Step 6. Interpret the result. The bottom row (except the remainder) gives the coefficients for the quotient polynomial. Since the dividend was of degree 2, the quotient will be of degree 1. The coefficients 2 and 5 correspond to:
[tex]$$2x + 5.$$[/tex]
Thus, the quotient in polynomial form 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 when performing synthetic division, we use the root of the divisor. Since
[tex]$$x + 1 = 0 \quad \Longrightarrow \quad x = -1,$$[/tex]
we will use [tex]$-1$[/tex] in our division steps.
Step 1. Write the coefficients of the dividend polynomial in order:
[tex]$$2, \quad 7, \quad 5.$$[/tex]
Step 2. Set up the synthetic division by writing [tex]$-1$[/tex] on the left and the coefficients to its right:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& & & \\
\end{array}
$$[/tex]
Step 3. Bring down the first coefficient (2):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & & \\
\end{array}
$$[/tex]
Step 4. Multiply the number you just brought down (2) by [tex]$-1$[/tex]:
[tex]$$2 \times (-1) = -2.$$[/tex]
Write this number under the next coefficient (7):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & -2 & \\
\end{array}
$$[/tex]
Then, add [tex]$7$[/tex] and [tex]$-2$[/tex]:
[tex]$$7 + (-2) = 5.$$[/tex]
Write the result (5) below the line:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & \\
\end{array}
$$[/tex]
Step 5. Multiply the new value (5) by [tex]$-1$[/tex]:
[tex]$$5 \times (-1) = -5.$$[/tex]
Place this number under the last coefficient (5):
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & -5 \\
\end{array}
$$[/tex]
Now, add the last column:
[tex]$$5 + (-5) = 0.$$[/tex]
Write the 0 as the remainder:
[tex]$$
\begin{array}{r|ccc}
-1 & 2 & 7 & 5 \\ \hline
& 2 & 5 & 0 \\
\end{array}
$$[/tex]
Step 6. Interpret the result. The bottom row (except the remainder) gives the coefficients for the quotient polynomial. Since the dividend was of degree 2, the quotient will be of degree 1. The coefficients 2 and 5 correspond to:
[tex]$$2x + 5.$$[/tex]
Thus, the quotient in polynomial form is
[tex]$$\boxed{2x+5}.$$[/tex]
