Final: No Training Wheels

When every term of a quadratic equation shares a number factor, take it out first. A number factor can never be zero, so it gives no solution of its own. Then factor the trinomial that is left and set each factor equal to zero.

For example, every term of $2x^{2}+10x+12 = 0$ is even, so write it as $2\left(x^{2}+5x+6\right) = 0$. The trinomial factors as $\left(x+2\right)\left(x+3\right)$, so the equation is $2\left(x+2\right)\left(x+3\right) = 0$. The factor 2 is never zero, so the solutions are $x = -2$ or $x = -3$. Taking out the 2 first makes the numbers in the factoring smaller.