To solve the problem [tex]\(\frac{x^2 + 4x + 7}{x + 3}\)[/tex] using long division, follow these steps:
1. Setup the long division:
[tex]\[
\begin{array}{r|rr}
x+3 & x^2+4x+7 \\
\end{array}
\][/tex]
2. Divide the first term:
- Take the leading term of the dividend [tex]\(x^2\)[/tex] and divide it by the leading term of the divisor [tex]\(x\)[/tex].
- [tex]\(\frac{x^2}{x} = x\)[/tex]
Write [tex]\(x\)[/tex] above the division bar.
[tex]\[
\begin{array}{r|rr}
& x \\
x+3 & x^2+4x+7 \\
\end{array}
\][/tex]
3. Multiply and subtract:
- Multiply [tex]\(x\)[/tex] by the divisor [tex]\(x + 3\)[/tex]:
[tex]\[
x \cdot (x + 3) = x^2 + 3x
\][/tex]
- Subtract this result from the original dividend:
[tex]\[
(x^2 + 4x + 7) - (x^2 + 3x) = (x^2 - x^2) + (4x - 3x) + 7 = x + 7
\][/tex]
Update the division process:
[tex]\[
\begin{array}{r|rr}
& x \\
x+3 & x^2+4x+7 \\
\hline
& x+7 \\
\end{array}
\][/tex]
4. Repeat the process:
- Now, take the leading term of the new polynomial [tex]\(x\)[/tex] and divide it by the leading term of the divisor [tex]\(x\)[/tex].
- [tex]\(\frac{x}{x} = 1\)[/tex]
Write [tex]\(1\)[/tex] above the division bar next to [tex]\(x\)[/tex].
[tex]\[
\begin{array}{r|rr}
& x+1 \\
x+3 & x^2+4x+7 \\
\hline
& x+7 \\
\end{array}
\][/tex]
5. Multiply and subtract again:
- Multiply [tex]\(1\)[/tex] by the divisor [tex]\(x + 3\)[/tex]:
[tex]\[
1 \cdot (x + 3) = x + 3
\][/tex]
- Subtract this result from the current polynomial:
[tex]\[
(x + 7) - (x + 3) = (x - x) + (7 - 3) = 4
\][/tex]
Update the division process:
[tex]\[
\begin{array}{r|rr}
& x+1 \\
x+3 & x^2+4x+7 \\
\hline
& x+7 \\
& x+3 \\
\hline
& 4 \\
\end{array}
\][/tex]
6. Conclusion:
- The quotient is the polynomial written above the division bar, which is [tex]\(x + 1\)[/tex].
- The remainder is the polynomial left over after the last subtraction, which is [tex]\(4\)[/tex].
Thus, the quotient is [tex]\(\boxed{x + 1}\)[/tex] and the remainder is [tex]\(\boxed{4}\)[/tex].
