Solve quadratic equations in one variable by factoring, using the zero-product property after writing the equation equal to zero (TN A1.A.REI.B.4).

What this topic is asking

Standard A1.A.REI.B.4 includes solving quadratics by factoring. The method rests on the zero-product property: if a product equals zero, at least one factor is zero. So you write the equation equal to zero, factor the quadratic, and set each factor to zero. The two resulting values are the solutions (the zeros or roots), which are also the $x$-intercepts of the parabola.

Why one side must be zero

The zero-product property is the engine, and it works only when a product equals zero. If $A \cdot B = 0$, then $A = 0$ or $B = 0$. But $A \cdot B = 21$ tells you nothing useful, because many factor pairs multiply to $21$. So $x^2 + 4x = 21$ must become $x^2 + 4x - 21 = 0$ before you factor.

Reading the solutions

Each factor $(x - r)$ contributes the solution $x = r$. So $(x - 5)$ gives $x = +5$ and $(x + 2)$ gives $x = -2$. The sign flips from the constant inside the factor, which is the single most common slip. If the quadratic factors as a perfect square, such as $(x - 3)^2 = 0$, there is one repeated solution $x = 3$ (a double root), and the parabola touches the $x$-axis there.

How TNReady examines this topic

• Multiple select. Select all solutions, with sign-flip and constant-value distractors.
• Numeric response. Solve and enter a solution (often the positive one).
• Multiple choice. Choose the solution set, or the zeros of a graphed parabola.

A clarifying idea is that solving by factoring and finding a function's zeros are the same task: the solutions of $x^2 - 3x - 10 = 0$ are exactly the $x$-intercepts of $y = x^2 - 3x - 10$. The quadratic-equations and quadratic-functions parts of the test meet here.

When factoring is and is not the right tool

Factoring is fastest when integer factors exist, and TNReady's multiple-choice quadratics usually do factor. But not every quadratic factors over the integers. A quick test is whether any integer pair multiplies to $ac$ (or to $c$ when the leading coefficient is $1$) and adds to $b$; if none does, the quadratic is prime over the integers, and you should switch to the quadratic formula or completing the square rather than hunting for factors that do not exist. Recognizing a prime trinomial quickly is itself a time-saving skill: it tells you to change methods instead of stalling. For a leading coefficient greater than $1$, use the $ac$ method (split the middle term and factor by grouping) before concluding it will not factor.

Special cases: GCF and difference of squares

Two structures simplify a factoring problem before you reach for a trinomial pair. First, factor out a common factor when every term shares one, which also exposes any hidden pattern: $3x^2 - 12x = 0$ becomes $3x(x - 4) = 0$, giving $x = 0$ or $x = 4$. Note that $x = 0$ is a genuine solution here, the factor $x$ contributes it, and it is easy to overlook. Second, a difference of squares factors directly: $x^2 - 25 = (x - 5)(x + 5) = 0$, so $x = \pm 5$, with no middle term to split.

Try this

Q1. Solve $x^2 - 7x + 12 = 0$. [1 point]

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

Q2. Solve $x^2 - 9 = 0$. [1 point]

• Cue. Difference of squares: $(x - 3)(x + 3) = 0$, so $x = 3$ or $x = -3$.