Think of two numbers that multiply to $-12$ and add up to $4$. Use those to factor $a^2 + 4a - 12$.

Once the quadratic is factored into two parentheses, remember: if a product is $0$, at least one factor must be $0$. What equations do you get from each factor, and which solution appears among the answer choices?

In Desmos, type y = x^2 + 4x - 12 to graph the function that corresponds to $p(a)$.

Look for the point where the graph crosses the x-axis (the x-intercept). Click that point to see its x-coordinate; that x-value is the solution of $p(a)=0$.

Compare the x-coordinate of the x-intercept you found in Desmos with the answer options $-8$, $-2$, $2$, and $6$, and select the matching value.

We are told $p(a)=a^2+4a-12$ and asked for which value of $a$ we have $p(a)=0$.

So we set the expression equal to zero:

Now we just need to solve this quadratic equation.

Look for two numbers that multiply to $-12$ and add to $4$.

Those numbers are $6$ and $-2$, since $6 \cdot (-2) = -12$ and $6 + (-2) = 4$.

So we can factor the quadratic:

Now the equation becomes:

If a product is zero, then at least one factor must be zero. So set each factor equal to zero:

Solve each equation:

• From $a + 6 = 0$, we get $a = -6$.
• From $a - 2 = 0$, we get $a = 2$.

The solutions of the equation are $a = -6$ and $a = 2$, but only $2$ appears in the answer choices, so the correct answer is $2$.