Factoring Polynomials

Why Factoring Runs Through SAT Advanced Math

Factoring rewrites a polynomial as a product of simpler pieces. It is rarely the whole SAT question — instead it is the move that unlocks the question. Given $x^2 - 7x + 12 = 0$ and asked for the sum of the solutions, factoring turns a scary equation into two tiny ones. Asked where a parabola crosses the $x$-axis, the factored form hands you the answers. Asked to simplify a rational expression, factoring both top and bottom exposes the common factor you cancel. Getting fast and accurate at the four core patterns pays off across the entire Advanced Math domain.

The First Move Is Always the GCF

Before anything else, pull out the greatest common factor — the largest factor shared by every term. In $6x^3 - 15x^2$ both terms share $3x^2$:

$6x^3 - 15x^2 = 3x^2(2x - 5).$

Removing the GCF early shrinks the numbers in every later step and often reveals a pattern that was hidden. Students who skip this step frequently conclude that a perfectly factorable expression "won't factor."

Pattern 1 — Difference of Squares

Any expression of the form $a^2 - b^2$ factors as

$a^2 - b^2 = (a - b)(a + b).$

So $x^2 - 49 = (x - 7)(x + 7)$ and $9x^2 - 25 = (3x - 5)(3x + 5)$. This pattern works only for a difference of two perfect squares. A sum of squares such as $x^2 + 49$ does not factor over the real numbers — that is a fact the SAT tests directly.

Pattern 2 — Simple Trinomials $x^2 + bx + c$

When the leading coefficient is $1$, find two numbers that multiply to $c$ and add to $b$. For $x^2 + 5x + 6$, you need a product of $6$ and a sum of $5$: those are $2$ and $3$, so

$x^2 + 5x + 6 = (x + 2)(x + 3).$

The signs are everything. If $c$ is positive and $b$ is negative, both numbers are negative (as in $x^2 - 9x + 20 = (x - 4)(x - 5)$). If $c$ is negative, the two numbers have opposite signs.

Pattern 3 — Trinomials With a Leading Coefficient (Grouping)

When $a \neq 1$, multiply $a \cdot c$, find two numbers that multiply to that product and add to $b$, split the middle term, and factor by grouping. For $2x^2 + 7x + 3$, compute $2 \cdot 3 = 6$ and find numbers multiplying to $6$ and adding to $7$ — namely $6$ and $1$:

$\begin{aligned} 2x^2 + 7x + 3 &= 2x^2 + 6x + x + 3 \\ &= 2x(x + 3) + 1(x + 3) \\ &= (2x + 1)(x + 3). \end{aligned}$

Grouping also handles any four-term polynomial: pair the terms, pull a common factor from each pair, and finish with the shared binomial.

Factor Completely — Then Read the Zeros

"Factor completely" means keep going until nothing else factors. Solve $3x^3 - 12x = 0$:

$\begin{aligned} 3x^3 - 12x &= 3x(x^2 - 4) \\ &= 3x(x - 2)(x + 2) = 0. \end{aligned}$

By the zero-product property, set each factor to zero: $x = 0$, $x = 2$, and $x = -2$. Stopping after the GCF would have hidden two of the three roots. The factored form is a gift: the zeros of $y = (x - 4)(x + 1)$ are exactly $x = 4$ and $x = -1$, because those inputs make one factor zero. Watch the sign flip — the factor $(x - 4)$ gives the root $x = +4$, and $(x + 1)$ gives $x = -1$.

Factoring to Simplify Expressions

The SAT also uses factoring to simplify rational expressions. To simplify

$\frac{x^2 - 9}{x^2 - x - 6},$

factor both parts: $x^2 - 9 = (x - 3)(x + 3)$ and $x^2 - x - 6 = (x - 3)(x + 2)$. The shared $(x - 3)$ cancels, leaving $\dfrac{x + 3}{x + 2}$ (with the understanding that $x \neq 3$). Recognizing that both numerator and denominator hide a common factor is the whole point of these problems.

The Recognition Checklist

1. GCF first — always.
2. Two terms? Check for difference of squares.
3. Three terms with $a = 1$? Product-and-sum trinomial.
4. Three terms with $a \neq 1$, or four terms? Grouping.
5. Anything left inside the parentheses? Keep factoring until nothing does.