Factor x^3 + 3x^2 - 4 completely.

Step-by-step solution

1. 1

   List the rational root candidates. The leading coefficient is $1$, so any rational root divides the constant $-4$:

   $\pm 1,\ \pm 2,\ \pm 4$

2. 2

   Test the small candidates. With $f(x)=x^3+3x^2-4$:

   $f(1)=1+3-4=0$

   So $x=1$ is a root and $(x-1)$ is a factor. (Note the missing $x$ term — its coefficient is $0$, which matters for the division.)

3. 3

   Divide by $(x-1)$ using synthetic division. Bring down the coefficients $1,\,3,\,0,\,-4$ with divisor $1$:

   $1\quad 4\quad 4\ \ |\ \ 0$

   The zero remainder confirms the root, and the quotient is $x^2+4x+4$.

4. 4

   Factor the quotient. It is a perfect square: $4=2^2$ and $2\cdot x\cdot 2=4x$:

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

5. 5

   Check by expansion and by the double root. $(x-1)(x+2)^2=(x-1)(x^2+4x+4)=x^3+3x^2-4$ $\checkmark$. Also $f(-2)=-8+12-4=0$, and since $-2$ is a double root the graph touches the axis there without crossing.