Factor polynomial expressions using common factors, the difference of two squares, perfect-square trinomials, and grouping (MA.912.AR.1.3).

What this topic is asking

Factoring reverses multiplication: MA.912.AR.1.3 asks you to rewrite a polynomial as a product of simpler factors. On the B.E.S.T. Algebra 1 EOC, factoring is the engine behind solving quadratics and rewriting expressions, so it appears both on its own and inside larger items. The methods are the GCF, trinomial factoring, the special patterns, and grouping.

Step one: the GCF

Before any other method, factor out the greatest common factor of all terms. For $6x^2 + 9x$, the GCF is $3x$, so $6x^2 + 9x = 3x(2x + 3)$. Pulling the GCF first often turns a hard-looking polynomial into an easy one, and skipping it leaves an "incomplete" factorization that the test marks wrong.

Factoring trinomials

For $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. For $x^2 + 5x + 6$, the numbers $2$ and $3$ multiply to $6$ and add to $5$, so it factors as $(x + 2)(x + 3)$. Sign rules:

• $c > 0$: both numbers share the sign of $b$.
• $c < 0$: the numbers have opposite signs, and the larger-magnitude one takes the sign of $b$.

The leading coefficient is not 1: the AC method

When $a \neq 1$, multiply $a \cdot c$, find two numbers multiplying to $ac$ and adding to $b$, split the middle term, and group.

The special patterns

Two patterns appear so often that you should recognize them on sight.

So $x^2 - 25 = (x + 5)(x - 5)$, and $x^2 + 6x + 9 = (x + 3)^2$ because $9 = 3^2$ and $6 = 2 \cdot 3$. A sum of squares, $x^2 + 25$, does not factor over the real numbers, a frequent distractor.

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

• Equation editor. Factor a polynomial completely and type the product.
• Multiple choice. Identify the complete factorization, with sign and incomplete-GCF distractors.
• Multiselect. Select all expressions that are factors of a given polynomial.

A clarifying idea: factoring and expanding are inverse operations, so the surest check is to multiply your factors back. If the product is not the original polynomial, the factorization is wrong, no matter how plausible it looks.

Why grouping works

Factoring by grouping rests on the distributive property applied twice. After splitting $3x^2 + 10x + 8$ into $3x^2 + 6x + 4x + 8$, the first pair shares $3x$ and the second shares $4$, giving $3x(x + 2) + 4(x + 2)$. Now $(x + 2)$ is itself a common factor of the two pieces, so it factors out, leaving $(3x + 4)(x + 2)$. The reason the two groups must share the same binomial is that the original split was chosen precisely so they would: the pair $6$ and $4$ was selected to multiply to $ac$, which guarantees the matching factor appears. Understanding this keeps you from picking a pair that adds to $b$ but does not multiply to $ac$.

Recognizing the special patterns saves time

Spotting a special pattern lets you skip the search for factor pairs entirely. A binomial is a difference of squares only when both terms are perfect squares and the sign between them is a minus: $x^2 - 49$, $9x^2 - 16$ (which is $(3x + 4)(3x - 4)$), and $25 - x^2$ all qualify. A trinomial is a perfect square when the first and last terms are perfect squares and the middle term is twice the product of their roots: in $x^2 + 6x + 9$, the roots are $x$ and $3$, and $2 \cdot x \cdot 3 = 6x$ matches the middle term, so it is $(x + 3)^2$. The EOC builds distractors out of near-misses, $x^2 + 9$ (a non-factorable sum of squares) or $x^2 + 5x + 9$ (where $5x \neq 2 \cdot 3x$), so confirming the middle term and the sign before you commit is the safeguard. When neither special pattern fits, fall back to the trinomial pair or the AC method.

Try this

Q1. Factor $4x^2 - 36$ completely. [2 points]

• Cue. GCF first: $4(x^2 - 9) = 4(x + 3)(x - 3)$.

Q2. Factor $x^2 - 8x + 16$. [1 point]

• Cue. Perfect square: $(x - 4)^2$.