Question

Solve the equation. Tell which solution method you used. \(24 x^{3}+18 x^{2}-168 x=0\)

Short answer

The solutions for the cubic equation \(24 x^{3}+18 x^{2}-168 x=0\) are \(x = 0\), \(x = \frac{7}{4}\), \(x = -4\). The solution method used was factoring.

Step-by-step solution

Simplifying the equation

Notice that each term in the equation \(24 x^{3}+18 x^{2}-168 x=0\) can be divided by 6. This gives us the simplified equation: \(4x^{3} + 3x^{2} - 28x = 0\).

Factoring out common terms

Now, notice that each term in the simplified equation has \(x\) as a common factor. Factor out \(x\) from each term to give: \(x(4x^{2} + 3x - 28) = 0\).

Factorizing the quadratic equation

We now have to solve the quadratic equation \(4x^{2} + 3x - 28\) for \(x\). This equation can be factorized to \((4x -7)(x + 4) = 0\). Thus the equation becomes \(x(4x - 7)(x + 4) = 0\).

Finding the solutions

Equation \(x(4x - 7)(x + 4) = 0\) will equal zero when either of \(x\), \(4x - 7\) or \(x + 4\) equals zero. The solutions for the equation are therefore \(x = 0\), \(x = \frac{7}{4}\), \(x = -4\).