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 (MA.912.AR.3.4).

What this topic is asking

MA.912.AR.3 asks you to solve a quadratic by factoring, using the zero-product property. The solutions are the zeros (x-intercepts) of the related parabola. On the B.E.S.T. Algebra 1 EOC this is the fastest method when the quadratic factors with integers, and it appears as equation-editor and multiple-choice items.

The zero-product property

This is the engine: a product is zero only when a factor is zero. It works only when one side is zero, which is why setting the equation to zero is the mandatory first step.

The method

1. Set to zero. Move all terms to one side: $x^2 + 3x = 10$ becomes $x^2 + 3x - 10 = 0$.
2. Factor. $(x + 5)(x - 2) = 0$.
3. Split. $x + 5 = 0$ or $x - 2 = 0$.
4. Solve. $x = -5$ or $x = 2$.

Solutions are the zeros

The solutions of $f(x) = 0$ are exactly the $x$-intercepts of $y = f(x)$, the points where the parabola crosses the $x$-axis. So "solve $x^2 - 7x + 12 = 0$" and "find the zeros of $f(x) = x^2 - 7x + 12$" are the same task. This link is why factoring is so useful for graphing.

How the B.E.S.T. EOC examines this topic

• Equation editor. Solve a factorable quadratic and enter both solutions.
• Multiple choice. Pick the solution set, with sign-reversal distractors.
• Multiselect. Select all the zeros of a function.

A clarifying idea: factoring reverses the multiplication that built the quadratic, and the zero-product property then turns one hard equation into two easy linear equations. That is the whole power of the method, splitting a degree-2 problem into two degree-1 problems.

Why setting to zero is mandatory

The first step is not a formality; the entire method collapses without it. The zero-product property only licenses the conclusion "one factor must be zero" when the product equals zero, because zero is the unique number with that property (no two nonzero numbers multiply to zero). If you factor $x^2 + 3x = 10$ as it stands into something times something equal to $10$, you learn nothing, since $10$ can be made many ways ($2 \times 5$, $1 \times 10$, $4 \times 2.5$, and so on), none of which pins down $x$. Only by writing $x^2 + 3x - 10 = 0$ and factoring the left side do you get a product equal to zero, where the property forces each factor to be considered. This is also why dividing both sides by $x$ is dangerous: it can erase the solution $x = 0$ before you ever apply the property.

Special cases: GCF and difference of squares

Not every factorable quadratic is a standard trinomial, and the EOC mixes in two shortcuts. When every term shares a factor, pull the GCF first: $2x^2 - 8x = 0$ becomes $2x(x - 4) = 0$, giving $x = 0$ or $x = 4$. Notice that factoring out $x$ (rather than dividing by it) preserves the $x = 0$ solution, which is why you never divide an equation by the variable. When the quadratic is a difference of squares with no middle term, factor it as a sum times a difference: $x^2 - 16 = (x + 4)(x - 4) = 0$, giving $x = \pm 4$. Recognizing these forms means you reach the zero-product step faster, without searching for a factor pair that the special pattern already hands you. After any factoring, the zero-product property finishes the solve identically.

Choosing factoring versus another method

Factoring is the fastest method when the quadratic factors over the integers, but it is a dead end when it does not. A quick test: compute $b^2 - 4ac$ mentally, or just try to find a factor pair multiplying to $ac$ and adding to $b$. If such a pair exists, factor; if the numbers are awkward (irrational roots, large primes), switch to the quadratic formula instead of forcing a factorization that will not appear. On the B.E.S.T. EOC, multiple-choice quadratics are usually designed to factor with small integers, so factoring is the efficient first attempt there, while equation-editor items asking for simplest radical form often signal that the formula is intended. Matching the method to the problem saves time you will want for the modeling items.

Try this

Q1. Solve $x^2 - 9 = 0$ by factoring. [1 point]

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

Q2. Solve $x^2 + 6x = 0$. [2 points]

• Cue. $x(x + 6) = 0$, so $x = 0$ or $x = -6$.