Answer :
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].
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].