Question
Find the zeroes of the polynomial equation 2 x cubed plus 6 x squared minus x minus 3 equals 0.
(1 point)
Responses
x equals negative 3, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction
Image with alt text: x equals negative 3 , Image with alt text: x equals Start Fraction Start Root 2 End Root over 2 End Fraction , Image with alt text: x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction
x equals negative 3, x equals negative 0.5, x = 0.5
Image with alt text: x equals negative 3 , Image with alt text: x equals negative 0.5 , x = 0.5
x equals negative Start Root 2 End Root, x equals Start Root 2 End Root, x = 3
Image with alt text: x equals negative Start Root 2 End Root , Image with alt text: x equals Start Root 2 End Root , x = 3
x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x = 3
(1 point)
Responses
x equals negative 3, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction
Image with alt text: x equals negative 3 , Image with alt text: x equals Start Fraction Start Root 2 End Root over 2 End Fraction , Image with alt text: x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction
x equals negative 3, x equals negative 0.5, x = 0.5
Image with alt text: x equals negative 3 , Image with alt text: x equals negative 0.5 , x = 0.5
x equals negative Start Root 2 End Root, x equals Start Root 2 End Root, x = 3
Image with alt text: x equals negative Start Root 2 End Root , Image with alt text: x equals Start Root 2 End Root , x = 3
x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x = 3
Answers
GPT-4o mini
To find the zeros of the polynomial equation \( 2x^3 + 6x^2 - x - 3 = 0 \), we can use the Rational Root Theorem, synthetic division, or numerical methods if necessary.
1. **Using the Rational Root Theorem**: According to the theorem, the possible rational roots could be the factors of the constant term divided by the factors of the leading coefficient.
The constant term is \(-3\) and the leading coefficient is \(2\). The possible roots are \(\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\).
2. **Testing possible rational roots**:
- Testing \(x = 1\):
\[
2(1)^3 + 6(1)^2 - (1) - 3 = 2 + 6 - 1 - 3 = 4 \quad (\text{not a root})
\]
- Testing \(x = -1\):
\[
2(-1)^3 + 6(-1)^2 - (-1) - 3 = -2 + 6 + 1 - 3 = 2 \quad (\text{not a root})
\]
- Testing \(x = 3\):
\[
2(3)^3 + 6(3)^2 - (3) - 3 = 54 + 54 - 3 - 3 = 102 \quad (\text{not a root})
\]
- Testing \(x = -3\):
\[
2(-3)^3 + 6(-3)^2 - (-3) - 3 = -54 + 54 + 3 - 3 = 0 \quad (\text{a root})
\]
3. **Factor the polynomial**: Since \(x = -3\) is a root, we can factor \(x + 3\) out of \(2x^3 + 6x^2 - x - 3\).
Using synthetic division to divide \(2x^3 + 6x^2 - x - 3\) by \(x + 3\):
\[
\begin{array}{r|rrrr}
-3 & 2 & 6 & -1 & -3 \\
& & -6 & 0 & 3 \\
\hline
& 2 & 0 & -1 & 0 \\
\end{array}
\]
The result is \(2x^2 - 1\). Thus,
\[
2x^3 + 6x^2 - x - 3 = (x + 3)(2x^2 - 1)
\]
4. **Setting the second factor to zero**:
\[
2x^2 - 1 = 0 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}
\]
5. **Combining the roots**: We have three roots:
- \(x = -3\)
- \(x = \frac{\sqrt{2}}{2}\)
- \(x = -\frac{\sqrt{2}}{2}\)
Thus, the correct response is:
**x equals negative 3, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction**.
1. **Using the Rational Root Theorem**: According to the theorem, the possible rational roots could be the factors of the constant term divided by the factors of the leading coefficient.
The constant term is \(-3\) and the leading coefficient is \(2\). The possible roots are \(\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\).
2. **Testing possible rational roots**:
- Testing \(x = 1\):
\[
2(1)^3 + 6(1)^2 - (1) - 3 = 2 + 6 - 1 - 3 = 4 \quad (\text{not a root})
\]
- Testing \(x = -1\):
\[
2(-1)^3 + 6(-1)^2 - (-1) - 3 = -2 + 6 + 1 - 3 = 2 \quad (\text{not a root})
\]
- Testing \(x = 3\):
\[
2(3)^3 + 6(3)^2 - (3) - 3 = 54 + 54 - 3 - 3 = 102 \quad (\text{not a root})
\]
- Testing \(x = -3\):
\[
2(-3)^3 + 6(-3)^2 - (-3) - 3 = -54 + 54 + 3 - 3 = 0 \quad (\text{a root})
\]
3. **Factor the polynomial**: Since \(x = -3\) is a root, we can factor \(x + 3\) out of \(2x^3 + 6x^2 - x - 3\).
Using synthetic division to divide \(2x^3 + 6x^2 - x - 3\) by \(x + 3\):
\[
\begin{array}{r|rrrr}
-3 & 2 & 6 & -1 & -3 \\
& & -6 & 0 & 3 \\
\hline
& 2 & 0 & -1 & 0 \\
\end{array}
\]
The result is \(2x^2 - 1\). Thus,
\[
2x^3 + 6x^2 - x - 3 = (x + 3)(2x^2 - 1)
\]
4. **Setting the second factor to zero**:
\[
2x^2 - 1 = 0 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}
\]
5. **Combining the roots**: We have three roots:
- \(x = -3\)
- \(x = \frac{\sqrt{2}}{2}\)
- \(x = -\frac{\sqrt{2}}{2}\)
Thus, the correct response is:
**x equals negative 3, x equals Start Fraction Start Root 2 End Root over 2 End Fraction, x equals negative Start Fraction Start Root 2 End Root over 2 End Fraction**.