Solve quadratic equations having real solutions by factoring, using the zero-product property, and relate the solutions to the zeros of the related quadratic function (TEKS A.8A).

What this topic is asking

Solving quadratics is the heart of the Quadratic Functions and Equations category. TEKS A.8A lists four methods, and factoring is the first and fastest when it works. The engine is the zero-product property, and the equation must be set to zero before you factor. STAAR also tests the connection between the solutions and the zeros (x-intercepts) of the related function.

Set to zero first

The zero-product property only works when one side is zero, so a quadratic must be written as $ax^2 + bx + c = 0$ before factoring.

Factoring while one side is nonzero is the single most common error: $x^2 + 5x = 24$ does not let you write $x(x + 5) = 24$ and conclude anything useful, because 24 is not zero.

Factor, then apply the zero-product property

Once at zero, factor (GCF first, then difference of squares or trinomial), then split into two linear equations.

For $x^2 + 5x - 24 = 0$: factor to $(x + 8)(x - 3) = 0$, so $x + 8 = 0$ or $x - 3 = 0$, giving $x = -8$ or $x = 3$.

Solutions are the zeros of the graph

The solutions of $ax^2 + bx + c = 0$ are exactly the $x$-intercepts (zeros) of the parabola $y = ax^2 + bx + c$, because an $x$-intercept is where $y = 0$. So solving the equation by factoring and reading the zeros off the graph give the same numbers. A quadratic with two real solutions crosses the $x$-axis twice; a perfect-square factorization gives a single (double) solution where the vertex touches the axis.

How STAAR examines this topic

• Multiple choice. Solve a factorable quadratic; the "signs reversed" answer is the standard distractor.
• Multiselect. Select all true statements about the solving process and the graph connection.
• Equation editor and number entry. Enter the solution(s), often as a set.

A clarifying idea is that factoring reverses the multiplication that built the quadratic, so the solutions are the values that make each factor vanish, which is why a solution is the opposite sign of the constant inside its factor.

When factoring is the right choice

Factoring is the fastest method, but only when the quadratic factors with integers. A quick way to decide is to check whether two integers multiply to $a \cdot c$ and add to $b$; if such a pair exists, factor, and if not, switch to the quadratic formula. On STAAR, the multiple-choice and multiselect quadratics are usually built to factor cleanly, so factoring is the expected first move, while the harder constructed items may force the formula. Recognizing a factorable quadratic on sight, especially the difference of squares $x^2 - 25 = 0$ and the simple trinomials, saves time you can spend on the multi-step problems.

Double roots and the difference of squares

Two special cases deserve attention. A perfect-square quadratic such as $x^2 - 6x + 9 = 0$ factors to $(x - 3)^2 = 0$, giving a single repeated solution $x = 3$, a double root where the parabola just touches the $x$-axis at its vertex. A difference of squares such as $x^2 - 16 = 0$ factors to $(x - 4)(x + 4) = 0$, giving the symmetric pair $x = \pm 4$. Both are common on the test precisely because they factor without the AC search, and recognizing them turns a solving problem into a one-line answer.

Try this

Q1. Solve $(x - 6)(x + 1) = 0$. [1 point]

• Cue. $x = 6$ or $x = -1$.

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

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