Solve the equation 4x^3 + 8x^3 + 4x^2 - 16x = 0.

Step-by-step solution

1. 1

   Combine like terms before anything else. The first two terms are both cubic:

   $4x^3 + 8x^3 = 12x^3 \quad\Longrightarrow\quad 12x^3 + 4x^2 - 16x = 0$

   Skipping this step and trying to factor four terms by grouping leads nowhere.

2. 2

   Pull out the greatest common factor. The coefficients $12, 4, 16$ share a factor $4$, and every term has at least one $x$:

   $4x\left(3x^2 + x - 4\right) = 0$

   This exposes $x = 0$ as a root and drops the remaining problem to a quadratic.

3. 3

   Factor the quadratic by splitting the middle term. For $3x^2 + x - 4$ we need two numbers with product $3 \times (-4) = -12$ and sum $+1$: those are $4$ and $-3$.

   $3x^2 + 4x - 3x - 4 = x(3x+4) - 1(3x+4) = (x-1)(3x+4)$

4. 4

   Set every factor to zero.

   $4x = 0 \Rightarrow x = 0, \qquad x - 1 = 0 \Rightarrow x = 1, \qquad 3x + 4 = 0 \Rightarrow x = -\frac{4}{3}$

5. 5

   Check each root in the original equation. $f(0) = 0$. $f(1) = 12 + 4 - 16 = 0$. $f(-\tfrac43) = 12\left(-\tfrac{64}{27}\right) + 4\left(\tfrac{16}{9}\right) + \tfrac{64}{3} = -\tfrac{256}{9} + \tfrac{64}{9} + \tfrac{192}{9} = 0$. All three are exact.