Solve the inequality 3x^2 - 2x - 1 >= 0.

Step-by-step solution

1. 1

   Use the AC method, since the leading coefficient is not 1. Multiply $a\cdot c=3\times(-1)=-3$ and look for two numbers with that product and with sum $b=-2$. They are $-3$ and $+1$:

   $(-3)(1)=-3\ \checkmark,\qquad -3+1=-2\ \checkmark$

2. 2

   Split the middle term and group.

   $3x^2-3x+x-1\ge0\qquad\Longrightarrow\qquad 3x(x-1)+1(x-1)\ge0$

   Both groups leave $(x-1)$, so it factors out:

   $(3x+1)(x-1)\ge0$

3. 3

   Find the critical points.

   $3x+1=0\Rightarrow x=-\frac13,\qquad x-1=0\Rightarrow x=1$

   These split the line into $\left(-\infty,-\tfrac13\right)$, $\left(-\tfrac13,1\right)$ and $(1,\infty)$.

4. 4

   Determine the sign pattern. The leading coefficient $3$ is positive, so the parabola opens upward: non-negative outside the roots and negative between them. Test points confirm it — at $x=-1$: $3+2-1=4>0$; at $x=0$: $-1<0$; at $x=2$: $12-4-1=7>0$.

5. 5

   Include the endpoints. The inequality is non-strict, so the roots, where the expression equals exactly $0$, belong to the solution set:

   $x\le-\frac13\qquad\text{or}\qquad x\ge1$

   In interval notation, $\left(-\infty,-\tfrac13\right]\cup[1,\infty)$.

6. 6

   Verify. The factorisation matches the original at every integer from $-30$ to $29$ ✓, and comparing the sign against the claimed solution set at $6001$ exact rational points on $[-10,10]$ gives agreement at every point ✓.