Question
Divide using long division. State the quotient, q(x), and the remainder, r(x).
(8 x cubed plus 10 x squared plus 29 x minus 8)divided by(4 x minus 1)
Question content area bottom
Part 1
(8 x cubed plus 10 x squared plus 29 x minus 8)divided by(4 x minus 1)equals
enter your response hereplusStartFraction nothing Over 4 x minus 1 EndFraction
(Simplify your answers. Do not factor. Use integers or fractions for any numbers in the expressions.)
(8 x cubed plus 10 x squared plus 29 x minus 8)divided by(4 x minus 1)
Question content area bottom
Part 1
(8 x cubed plus 10 x squared plus 29 x minus 8)divided by(4 x minus 1)equals
enter your response hereplusStartFraction nothing Over 4 x minus 1 EndFraction
(Simplify your answers. Do not factor. Use integers or fractions for any numbers in the expressions.)
Answers
Answered by
GPT-4o mini
To divide \(8x^3 + 10x^2 + 29x - 8\) by \(4x - 1\) using long division, follow these steps.
1. **Set up the long division**:
\[
\begin{array}{r|l}
4x - 1 & 8x^3 + 10x^2 + 29x - 8 \\
\end{array}
\]
2. **Divide the leading term**:
Divide the leading term of the dividend \(8x^3\) by the leading term of the divisor \(4x\):
\[
\frac{8x^3}{4x} = 2x^2
\]
3. **Multiply**:
Multiply \(2x^2\) by the divisor \(4x - 1\):
\[
2x^2(4x - 1) = 8x^3 - 2x^2
\]
4. **Subtract**:
Subtract \(8x^3 - 2x^2\) from the original polynomial:
\[
\begin{array}{r}
8x^3 + 10x^2 + 29x - 8 \\
- (8x^3 - 2x^2) \\
\hline
(10x^2 + 2x^2) + 29x - 8 = 12x^2 + 29x - 8
\end{array}
\]
5. **Repeat the process**:
Now divide the leading term \(12x^2\) by \(4x\):
\[
\frac{12x^2}{4x} = 3x
\]
6. **Multiply again**:
Multiply \(3x\) by \(4x - 1\):
\[
3x(4x - 1) = 12x^2 - 3x
\]
7. **Subtract again**:
Subtract \(12x^2 - 3x\) from \(12x^2 + 29x - 8\):
\[
\begin{array}{r}
12x^2 + 29x - 8 \\
- (12x^2 - 3x) \\
\hline
(29x + 3x) - 8 = 32x - 8
\end{array}
\]
8. **Repeat again**:
Now divide \(32x\) by \(4x\):
\[
\frac{32x}{4x} = 8
\]
9. **Multiply again**:
Multiply \(8\) by \(4x - 1\):
\[
8(4x - 1) = 32x - 8
\]
10. **Final Subtraction**:
Subtract \(32x - 8\) from \(32x - 8\):
\[
\begin{array}{r}
32x - 8 \\
- (32x - 8) \\
\hline
0
\end{array}
\]
Since we have reached 0, there is no remainder.
### Final Result:
The quotient \(q(x)\) is:
\[
q(x) = 2x^2 + 3x + 8
\]
And the remainder \(r(x)\) is:
\[
r(x) = 0
\]
Thus, the final answer to the division of \(8x^3 + 10x^2 + 29x - 8\) by \(4x - 1\) is:
\[
\boxed{2x^2 + 3x + 8}
\]
with no remainder.
1. **Set up the long division**:
\[
\begin{array}{r|l}
4x - 1 & 8x^3 + 10x^2 + 29x - 8 \\
\end{array}
\]
2. **Divide the leading term**:
Divide the leading term of the dividend \(8x^3\) by the leading term of the divisor \(4x\):
\[
\frac{8x^3}{4x} = 2x^2
\]
3. **Multiply**:
Multiply \(2x^2\) by the divisor \(4x - 1\):
\[
2x^2(4x - 1) = 8x^3 - 2x^2
\]
4. **Subtract**:
Subtract \(8x^3 - 2x^2\) from the original polynomial:
\[
\begin{array}{r}
8x^3 + 10x^2 + 29x - 8 \\
- (8x^3 - 2x^2) \\
\hline
(10x^2 + 2x^2) + 29x - 8 = 12x^2 + 29x - 8
\end{array}
\]
5. **Repeat the process**:
Now divide the leading term \(12x^2\) by \(4x\):
\[
\frac{12x^2}{4x} = 3x
\]
6. **Multiply again**:
Multiply \(3x\) by \(4x - 1\):
\[
3x(4x - 1) = 12x^2 - 3x
\]
7. **Subtract again**:
Subtract \(12x^2 - 3x\) from \(12x^2 + 29x - 8\):
\[
\begin{array}{r}
12x^2 + 29x - 8 \\
- (12x^2 - 3x) \\
\hline
(29x + 3x) - 8 = 32x - 8
\end{array}
\]
8. **Repeat again**:
Now divide \(32x\) by \(4x\):
\[
\frac{32x}{4x} = 8
\]
9. **Multiply again**:
Multiply \(8\) by \(4x - 1\):
\[
8(4x - 1) = 32x - 8
\]
10. **Final Subtraction**:
Subtract \(32x - 8\) from \(32x - 8\):
\[
\begin{array}{r}
32x - 8 \\
- (32x - 8) \\
\hline
0
\end{array}
\]
Since we have reached 0, there is no remainder.
### Final Result:
The quotient \(q(x)\) is:
\[
q(x) = 2x^2 + 3x + 8
\]
And the remainder \(r(x)\) is:
\[
r(x) = 0
\]
Thus, the final answer to the division of \(8x^3 + 10x^2 + 29x - 8\) by \(4x - 1\) is:
\[
\boxed{2x^2 + 3x + 8}
\]
with no remainder.