Question
Use the long division method to find the result when 3, x, cubed, plus, 10, x, squared, plus, 15, x, plus, 143x
3
+10x
2
+15x+14 is divided by x, plus, 2x+2.
Solve the problem just like the example below
⬇️
Use the long division method to find the result when
4, x, cubed, plus, 5, x, squared, plus, 5, x, plus, 1
4x
3
+5x
2
+5x+1 is divided by
4, x, plus, 1
4x+1.
Use Long Division:
Use Long Division:
x, squared
x
2
plus, x
+x
plus, 1
+1
4, x
4x
plus, 1
+1
right parenthesis
)
4, x, cubed
4x
3
plus, 5, x, squared
+5x
2
plus, 5, x
+5x
plus, 1
+1
start superscript, minus, end superscript, 4, x, cubed
−
4x
3
plus, start superscript, minus, end superscript, x, squared
+
−
x
2
4, x, squared
4x
2
plus, 5, x
+5x
plus, 1
+1
start superscript, minus, end superscript, 4, x, squared
−
4x
2
plus, start superscript, minus, end superscript, x
+
−
x
4, x
4x
plus, 1
+1
start superscript, minus, end superscript, 4, x
−
4x
plus, start superscript, minus, end superscript, 1
+
−
1
0
0
Answer: , x, squared, plus, x, plus, 1
Answer: x
2
+x+1
3
+10x
2
+15x+14 is divided by x, plus, 2x+2.
Solve the problem just like the example below
⬇️
Use the long division method to find the result when
4, x, cubed, plus, 5, x, squared, plus, 5, x, plus, 1
4x
3
+5x
2
+5x+1 is divided by
4, x, plus, 1
4x+1.
Use Long Division:
Use Long Division:
x, squared
x
2
plus, x
+x
plus, 1
+1
4, x
4x
plus, 1
+1
right parenthesis
)
4, x, cubed
4x
3
plus, 5, x, squared
+5x
2
plus, 5, x
+5x
plus, 1
+1
start superscript, minus, end superscript, 4, x, cubed
−
4x
3
plus, start superscript, minus, end superscript, x, squared
+
−
x
2
4, x, squared
4x
2
plus, 5, x
+5x
plus, 1
+1
start superscript, minus, end superscript, 4, x, squared
−
4x
2
plus, start superscript, minus, end superscript, x
+
−
x
4, x
4x
plus, 1
+1
start superscript, minus, end superscript, 4, x
−
4x
plus, start superscript, minus, end superscript, 1
+
−
1
0
0
Answer: , x, squared, plus, x, plus, 1
Answer: x
2
+x+1
Answers
GPT-4o mini
To find the result of dividing \( 3x^3 + 10x^2 + 15x + 14 \) by \( x + 2 \) using the long division method, we will follow the same steps demonstrated in your example.
### Use Long Division:
1. **Set up the Division**:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 3x^3 + 10x^2 + 15x + 14 \\
\end{array}
\]
2. **Divide the first term**:
- Divide \( 3x^3 \) by \( x \) to get \( 3x^2 \).
3. **Multiply and subtract**:
- Multiply \( 3x^2 \) by \( (x + 2) \):
\[
3x^2(x + 2) = 3x^3 + 6x^2
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 3x^3 + 10x^2 + 15x + 14 \\
- & (3x^3 + 6x^2) \\
\hline
& 0 + 4x^2 + 15x + 14 \\
\end{array}
\]
4. **Divide the next term**:
- Now divide \( 4x^2 \) by \( x \) to get \( 4x \).
5. **Multiply and subtract again**:
- Multiply \( 4x \) by \( (x + 2) \):
\[
4x(x + 2) = 4x^2 + 8x
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 4x^2 + 15x + 14 \\
- & (4x^2 + 8x) \\
\hline
& 0 + 7x + 14 \\
\end{array}
\]
6. **Divide the next term**:
- Now divide \( 7x \) by \( x \) to get \( 7 \).
7. **Multiply and subtract again**:
- Multiply \( 7 \) by \( (x + 2) \):
\[
7(x + 2) = 7x + 14
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 7x + 14 \\
- & (7x + 14) \\
\hline
& 0 \\
\end{array}
\]
### Final Result
Since we have reached a remainder of 0, the division is complete. Combining our results, we have:
\[
3x^3 + 10x^2 + 15x + 14 \div (x + 2) = 3x^2 + 4x + 7
\]
**Answer:** \( 3x^2 + 4x + 7 \)
### Use Long Division:
1. **Set up the Division**:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 3x^3 + 10x^2 + 15x + 14 \\
\end{array}
\]
2. **Divide the first term**:
- Divide \( 3x^3 \) by \( x \) to get \( 3x^2 \).
3. **Multiply and subtract**:
- Multiply \( 3x^2 \) by \( (x + 2) \):
\[
3x^2(x + 2) = 3x^3 + 6x^2
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 3x^3 + 10x^2 + 15x + 14 \\
- & (3x^3 + 6x^2) \\
\hline
& 0 + 4x^2 + 15x + 14 \\
\end{array}
\]
4. **Divide the next term**:
- Now divide \( 4x^2 \) by \( x \) to get \( 4x \).
5. **Multiply and subtract again**:
- Multiply \( 4x \) by \( (x + 2) \):
\[
4x(x + 2) = 4x^2 + 8x
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 4x^2 + 15x + 14 \\
- & (4x^2 + 8x) \\
\hline
& 0 + 7x + 14 \\
\end{array}
\]
6. **Divide the next term**:
- Now divide \( 7x \) by \( x \) to get \( 7 \).
7. **Multiply and subtract again**:
- Multiply \( 7 \) by \( (x + 2) \):
\[
7(x + 2) = 7x + 14
\]
- Now subtract:
\[
\begin{array}{r|l}
& x + 2 \\
\hline
& 7x + 14 \\
- & (7x + 14) \\
\hline
& 0 \\
\end{array}
\]
### Final Result
Since we have reached a remainder of 0, the division is complete. Combining our results, we have:
\[
3x^3 + 10x^2 + 15x + 14 \div (x + 2) = 3x^2 + 4x + 7
\]
**Answer:** \( 3x^2 + 4x + 7 \)