Question

1. $2x^{3}+24x^{2}-56x = 0$

Explanation:

Step1: Factor out the greatest common factor

First, we can see that each term in the equation \(2x^{3}+24x^{2}-56x = 0\) has a common factor of \(2x\). So we factor out \(2x\):
\(2x(x^{2}+12x - 28)=0\)

Step2: Set each factor equal to zero

According to the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have two cases:
Case 1: \(2x=0\)
Solving for \(x\), we divide both sides by 2: \(x = 0\)

Case 2: \(x^{2}+12x - 28=0\)
We can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 12\), and \(c=-28\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(12)^{2}-4\times1\times(-28)=144 + 112=256\)
Then, \(x=\frac{-12\pm\sqrt{256}}{2\times1}=\frac{-12\pm16}{2}\)

We have two sub - cases for this:
Sub - case 1: \(x=\frac{-12 + 16}{2}=\frac{4}{2}=2\)
Sub - case 2: \(x=\frac{-12-16}{2}=\frac{-28}{2}=-14\)

Answer:

The solutions of the equation \(2x^{3}+24x^{2}-56x = 0\) are \(x = 0\), \(x = 2\) and \(x=-14\)