Solve quadratic equations in one variable by factoring and applying the zero product property, and interpret the solutions as the zeros of the related function (A.EI.6).

What this topic is asking

A.EI.6 asks you to solve quadratic equations by factoring. On the Virginia Algebra I SOL these are Equations and Inequalities items: factor a quadratic set equal to zero, apply the zero product property, and state the solutions (which are the zeros of the related function). They appear as fill-in-the-blank (type both solutions), multiple choice, and drag-and-drop.

Step 1: set the equation to zero

Factoring only solves a quadratic when one side is zero, because the method relies on a product being zero. So move every term to one side first:

Solving $x^2 + 5x = 6$ by factoring the left side as $x(x + 5) = 6$ and setting factors to $6$ does not work, the zero product property needs a zero.

Step 2: the zero product property

The key fact is:

If $A \cdot B = 0$, then $A = 0$ or $B = 0$ (or both).

A product of real numbers is zero only when at least one factor is zero. So once the quadratic is factored and equals zero, you split it into one linear equation per factor.

Reading solutions from factors

The solutions come from setting each factor to zero, so the signs flip relative to the numbers in the factors. The factor $(x + 5)$ gives $x = -5$, and $(x - 3)$ gives $x = 3$. A very common slip is reading $(x + 5)(x - 3) = 0$ as solutions $5$ and $-3$; it is the opposite. If a factor has a coefficient, like $(2x - 1)$, solve it fully: $2x - 1 = 0$ gives $x = \frac{1}{2}$.

Special cases

• Difference of squares. $x^2 - 16 = 0$ factors as $(x + 4)(x - 4) = 0$, so $x = \pm 4$.
• Common factor. $2x^2 - 8x = 0$ factors as $2x(x - 4) = 0$, so $x = 0$ or $x = 4$. Do not divide both sides by $x$, that loses the $x = 0$ solution.
• Perfect-square trinomial. $x^2 - 6x + 9 = 0$ is $(x - 3)^2 = 0$, a double root at $x = 3$ (one repeated solution).

Why the zeros are the x-intercepts

The solutions of $ax^2 + bx + c = 0$ are exactly the $x$-intercepts of the parabola $y = ax^2 + bx + c$, and understanding why ties this topic to graphing. An $x$-intercept is a point where the graph crosses the $x$-axis, which means $y = 0$ there. So asking "where does the parabola cross the $x$-axis?" is the same as asking "for which $x$ is $ax^2 + bx + c = 0$?", which is the equation you are solving. Each factor of the quadratic corresponds to one intercept: $(x + 5)(x - 3) = 0$ tells you the parabola crosses at $x = -5$ and $x = 3$. This is why a quadratic with two real roots has a parabola crossing the axis twice, a double root touches the axis once (at the vertex), and a quadratic that does not factor over the reals has a parabola that never reaches the axis. Solving by factoring is, geometrically, locating where the curve meets the $x$-axis.

How the SOL examines this topic

• Fill-in-the-blank. Solve a factorable quadratic and type both solutions.
• Multiple choice. Pick the solutions, with distractors that keep the wrong signs or only one root.
• Drag-and-drop. Order the steps, or match a quadratic to its solution set.

Try this

Q1. Solve $x^2 - 7x + 12 = 0$. [2 points]

• Cue. $(x - 3)(x - 4) = 0$, so $x = 3$ or $x = 4$.

Q2. Solve $x^2 + 4x = 0$. [1 point]

• Cue. $x(x + 4) = 0$, so $x = 0$ or $x = -4$.