Factor 3x^3 + 7x^2 - 4 completely.

Step-by-step solution

1. 1

   Build the candidate list from both end coefficients. A rational root $p/q$ needs $p\mid 4$ and $q\mid 3$:

   $\pm 1,\ \pm 2,\ \pm 4,\ \pm\tfrac13,\ \pm\tfrac23,\ \pm\tfrac43$

2. 2

   Test the easy integers. With $f(x)=3x^3+7x^2-4$:

   $f(-1)=-3+7-4=0$

   so $x=-1$ is a root and $(x+1)$ is a factor.

3. 3

   Divide by $(x+1)$ with synthetic division. Coefficients $3,\,7,\,0,\,-4$ and divisor $-1$:

   $3\quad 4\quad -4\ \ |\ \ 0$

   giving the quotient $3x^2+4x-4$ with zero remainder.

4. 4

   Factor the quadratic by the AC method. $ac=3\cdot(-4)=-12$, and we need a pair summing to $4$: that is $6$ and $-2$:

   $3x^2+6x-2x-4=3x(x+2)-2(x+2)=(3x-2)(x+2)$

5. 5

   Assemble and verify.

   $3x^3+7x^2-4=(x+1)(x+2)(3x-2)$

   Expanding: $(x+1)(x+2)=x^2+3x+2$, and $(x^2+3x+2)(3x-2)=3x^3-2x^2+9x^2-6x+6x-4=3x^3+7x^2-4$ $\checkmark$. The three roots are $-1$, $-2$ and $\tfrac23$; note $f\left(\tfrac23\right)=3\cdot\tfrac{8}{27}+7\cdot\tfrac49-4=\tfrac89+\tfrac{28}{9}-4=0$ $\checkmark$.