Question
Find a step-by-step solution to the equation x^3 + 2x^2 - 7x - 12 = 0.
Answers
GPT-4o mini
To solve the cubic equation \( x^3 + 2x^2 - 7x - 12 = 0 \), we can use synthetic division to test for rational roots. According to the Rational Root Theorem, the possible rational roots are the factors of the constant term divided by the factors of the leading coefficient.
In this case, the constant term is -12, and the leading coefficient is 1. The factors of -12 are:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12
\]
We'll start testing these possible roots:
1. **Test \( x = 2 \):**
\[
2^3 + 2(2^2) - 7(2) - 12 = 8 + 8 - 14 - 12 = -10 \quad \text{(not a root)}
\]
2. **Test \( x = -2 \):**
\[
(-2)^3 + 2(-2)^2 - 7(-2) - 12 = -8 + 8 + 14 - 12 = 2 \quad \text{(not a root)}
\]
3. **Test \( x = -3 \):**
\[
(-3)^3 + 2(-3)^2 - 7(-3) - 12 = -27 + 18 + 21 - 12 = 0 \quad \text{(is a root)}
\]
Since \( x = -3 \) is a root, we can now perform synthetic division to factor the polynomial by \( x + 3 \).
**Synthetic Division:**
\[
\begin{array}{r|rrrr}
-3 & 1 & 2 & -7 & -12 \\
& & -3 & 3 & 12 \\
\hline
& 1 & -1 & -4 & 0 \\
\end{array}
\]
The result of the synthetic division is \( x^2 - x - 4 \).
Now we have the factored form of the cubic polynomial:
\[
x^3 + 2x^2 - 7x - 12 = (x + 3)(x^2 - x - 4)
\]
Next, we solve the quadratic equation \( x^2 - x - 4 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting \( a = 1, b = -1, c = -4 \):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}
\]
Thus, the roots of the equation \( x^2 - x - 4 = 0 \) are:
\[
x = \frac{1 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{17}}{2}
\]
Therefore, the solutions to the original equation \( x^3 + 2x^2 - 7x - 12 = 0 \) are:
\[
\boxed{-3}, \quad \frac{1 + \sqrt{17}}{2}, \quad \frac{1 - \sqrt{17}}{2}
\]
In this case, the constant term is -12, and the leading coefficient is 1. The factors of -12 are:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12
\]
We'll start testing these possible roots:
1. **Test \( x = 2 \):**
\[
2^3 + 2(2^2) - 7(2) - 12 = 8 + 8 - 14 - 12 = -10 \quad \text{(not a root)}
\]
2. **Test \( x = -2 \):**
\[
(-2)^3 + 2(-2)^2 - 7(-2) - 12 = -8 + 8 + 14 - 12 = 2 \quad \text{(not a root)}
\]
3. **Test \( x = -3 \):**
\[
(-3)^3 + 2(-3)^2 - 7(-3) - 12 = -27 + 18 + 21 - 12 = 0 \quad \text{(is a root)}
\]
Since \( x = -3 \) is a root, we can now perform synthetic division to factor the polynomial by \( x + 3 \).
**Synthetic Division:**
\[
\begin{array}{r|rrrr}
-3 & 1 & 2 & -7 & -12 \\
& & -3 & 3 & 12 \\
\hline
& 1 & -1 & -4 & 0 \\
\end{array}
\]
The result of the synthetic division is \( x^2 - x - 4 \).
Now we have the factored form of the cubic polynomial:
\[
x^3 + 2x^2 - 7x - 12 = (x + 3)(x^2 - x - 4)
\]
Next, we solve the quadratic equation \( x^2 - x - 4 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substituting \( a = 1, b = -1, c = -4 \):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}
\]
Thus, the roots of the equation \( x^2 - x - 4 = 0 \) are:
\[
x = \frac{1 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{17}}{2}
\]
Therefore, the solutions to the original equation \( x^3 + 2x^2 - 7x - 12 = 0 \) are:
\[
\boxed{-3}, \quad \frac{1 + \sqrt{17}}{2}, \quad \frac{1 - \sqrt{17}}{2}
\]