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

Step-by-step solution

1. 1

   Factor out the common x. Every term carries an $x$:

   $x\left(3x^3 + 8x^2 - 4x - 16\right) = 0$

   so $x = 0$ is one root immediately, and the rest come from the cubic.

2. 2

   Do not trust factoring by grouping here. The tempting move is $\left(3x^3 + 8x^2\right) + \left(-4x - 16\right) = x^2(3x+8) - 4(x+4)$. The two brackets are $3x+8$ and $x+4$ — they are not the same, so grouping does not close. Writing the second as $-4(3x+8)$ would mean $-12x - 32$, not $-4x - 16$; that slip is the usual source of the wrong answer $x = -\tfrac83,\,\pm 2$ for this equation.

3. 3

   Find a rational root of the cubic instead. Candidates are $\pm\tfrac{d}{e}$ with $d \mid 16$, $e \mid 3$. Testing $x = -2$:

   $3(-8) + 8(4) - 4(-2) - 16 = -24 + 32 + 8 - 16 = 0$

   so $(x+2)$ divides the cubic.

4. 4

   Divide and factor the quotient. Dividing $3x^3 + 8x^2 - 4x - 16$ by $(x+2)$ gives $3x^2 + 2x - 8$, which factors as $(3x - 4)(x + 2)$. Hence

   $3x^3 + 8x^2 - 4x - 16 = (x+2)^2(3x-4)$

   and the full factorisation is

   $3x^4 + 8x^3 - 4x^2 - 16x = x(x+2)^2(3x-4)$

5. 5

   Read off the roots. Setting each factor to zero:

   $x = 0, \qquad x = -2 \ \text{(double root)}, \qquad x = \frac{4}{3}$

6. 6

   Verify. Expanding $x(x+2)^2(3x-4)$ returns the coefficients $3,\ 8,\ -4,\ -16,\ 0$ exactly. Direct substitution: $f(0) = 0$, $f(-2) = 48 - 64 - 16 + 32 = 0$, $f(4/3) = 0$ to within $10^{-14}$. Because $x = -2$ is a double root, the graph touches the axis there without crossing.