Choose and produce equivalent forms of an expression, factoring a quadratic and using the structure to reveal zeros, a maximum or minimum, or other properties (LA A1: A-SSE.B.3).

What this topic is asking

Standard A1: A-SSE.B.3 asks you to rewrite an expression in an equivalent form so its structure reveals a useful property, most often factoring a quadratic to expose its zeros, or producing a form that shows a maximum or minimum. On LEAP 2025 these are Type I Major Content items: factor a trinomial, recognize a special product, or pick the equivalent form. Factoring is also a core no-calculator skill, so expect it in Session 1a.

Factoring a simple trinomial

For $x^2 + bx + c$ (leading coefficient $1$), find two numbers that multiply to $c$ and add to $b$. The signs follow $c$ and $b$.

A negative constant means the two numbers have opposite signs; a positive constant with a negative middle term means both are negative.

Always factor out the GCF first

Before any other move, remove the greatest common factor of all terms. For $2x^2 + 12x + 16$, the GCF is $2$: rewrite as $2(x^2 + 6x + 8) = 2(x + 2)(x + 4)$. Skipping the GCF leaves an unfinished factorization and is a frequent lost point, because "factor completely" means no common factor remains.

The special patterns

These patterns are not on the reference sheet, so memorize them. Recognizing a difference of squares or a perfect square lets you factor in one step instead of searching for factor pairs. A difference of squares always has the form (something squared) minus (something squared) with no middle term: $x^2 - 25$, $9x^2 - 16$, or $x^2 - 1$. A perfect-square trinomial has a middle term equal to twice the product of the two square roots: in $x^2 + 10x + 25$ the roots are $x$ and $5$, and $2 \cdot x \cdot 5 = 10x$ is exactly the middle term, which confirms the pattern.

Factoring out a common variable

When every term shares a variable, factor it out along with any numerical GCF. For $x^2 - 5x$, both terms contain an $x$, so $x^2 - 5x = x(x - 5)$, with zeros at $x = 0$ and $x = 5$. This is the GCF idea applied to a variable, and it is a frequent setup in the quadratics module, where a missing constant term means the GCF is the fastest route to the zeros. Do not divide the $x$ away, that would discard the $x = 0$ solution; factor it out so the zero is preserved.

How LEAP examines this topic

• Equation response. Factor a quadratic completely and type the factored form.
• Multiple choice. Pick the equivalent form, or identify which form reveals the zeros.
• Drag and drop. Build the factors from tiles, or match a quadratic to its factored form.

Why factored form reveals the zeros

Factored form is prized because of the zero product property: a product is zero exactly when one of its factors is zero. So if $x^2 + bx + c = (x - p)(x - q)$, the expression equals zero precisely at $x = p$ and $x = q$, which are the zeros of the related function and the $x$-intercepts of its parabola. Standard form $ax^2 + bx + c$ hides those values, while factored form displays them. This is the heart of A-SSE.B.3, choosing the form that exposes the property you want: factored form for zeros, and (in the completing-the-square topic) vertex form for the maximum or minimum. The same expression carries the same information in every form, but each form makes a different feature easy to read.

Try this

Q1. Factor $x^2 + 8x + 16$. [2 points]

• Cue. Perfect square: $(x + 4)^2$.

Q2. Factor $3x^2 - 27$ completely. [2 points]

• Cue. GCF first: $3(x^2 - 9) = 3(x + 3)(x - 3)$.