x3-6x2-19x-12=0
To solve the cubic equation \( x^3 - 6x^2 - 19x - 12 = 0 \), we can use the Rational Root Theorem, synthetic division, or factoring if possible.
- Finding rational roots: According to the Rational Root Theorem, potential rational roots are factors of the constant term (\(-12\)) divided by factors of the leading coefficient (\(1\)). The factors of \(-12\) are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\).
We can test these values to see if any of them are roots of the equation.
- Testing \(x = -1\): \[ (-1)^3 - 6(-1)^2 - 19(-1) - 12 = -1 - 6 + 19 - 12 = 0 \] So, \(x = -1\) is a root.
- Synthetic Division: Now that we know \(x + 1\) is a factor, we can perform synthetic division to divide the polynomial by \(x + 1\).
\[ \begin{array}{r|rrrr} -1 & 1 & -6 & -19 & -12 \\ & & -1 & 7 & 12 \\ \hline & 1 & -7 & -12 & 0 \\ \end{array} \]
This gives us: \[ x^3 - 6x^2 - 19x - 12 = (x + 1)(x^2 - 7x - 12) \]
- Factoring the quadratic \(x^2 - 7x - 12\): Now, we need to factor \(x^2 - 7x - 12\): We need two numbers that multiply to \(-12\) and add to \(-7\). The numbers \(-8\) and \(1\) work because \(-8 \times 1 = -12\) and \(-8 + 1 = -7\).
Thus, \[ x^2 - 7x - 12 = (x - 8)(x + 1) \]
Complete factorization: Putting it all together, we have: \[ x^3 - 6x^2 - 19x - 12 = (x + 1)(x - 8)(x + 1) = (x + 1)^2(x - 8) \]
Finding the roots: Setting each factor to zero gives: \[ (x + 1)^2 = 0 \quad \Rightarrow \quad x = -1 \quad (\text{double root}) \] \[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \]
Thus, the solutions to the equation \(x^3 - 6x^2 - 19x - 12 = 0\) are: \[ x = -1 \quad (\text{double root}), \quad x = 8. \]
