Factoring

Why bother factoring?

Three reasons:

1. Zeros. The solutions to $p(x) = 0$ are the $x$-values that make $p(x) = 0$. If $p(x) = (x-2)(x+5)$, then the zero-product property says the only solutions are $x = 2$ and $x = -5$. We'll use this immediately.

2. Simplification. Rational expressions (next lesson) need factored numerators and denominators to cancel common factors.

3. Structure. Factored form reveals what a polynomial is built from, the way prime factorization reveals what a number is built from.

The zero-product property: a theorem

If $a \cdot b = 0$, then $a = 0$ or $b = 0$ (or both).

This is provable from the properties of real numbers: if $a \ne 0$, multiply both sides by $\frac{1}{a}$ (which exists because $a \ne 0$), giving $b = 0$. Done. This theorem is true for real numbers; it fails in "rings with zero divisors" — an algebraic structure you might meet later. For now, it's the engine of everything.

Step 1: GCF first, always

Before doing anything fancy, pull out the greatest common factor of all terms.

$6x^3 - 9x^2 + 3x = 3x(2x^2 - 3x + 1).$

Why? Because every subsequent step is easier on a smaller polynomial. Skipping GCF is the source of half the errors in factoring.

Step 2: trinomials $x^2 + bx + c$

To factor $x^2 + bx + c$: find two integers $p$ and $q$ such that $p \cdot q = c$ and $p + q = b$. Then:

$x^2 + bx + c = (x+p)(x+q).$

Why? FOIL confirms: $(x+p)(x+q) = x^2 + (p+q)x + pq = x^2 + bx + c$. We're undoing FOIL.

Example: $x^2 + 5x + 6$. Need $pq = 6$, $p+q = 5$. Pairs for 6: $(1,6), (2,3), (-1,-6), (-2,-3)$. The pair $(2,3)$ sums to 5. So $x^2+5x+6 = (x+2)(x+3)$.

Example: $x^2 - 4x - 12$. Need $pq = -12$, $p+q = -4$. The pair $(2, -6)$: $2 \cdot (-6) = -12$, $2+(-6) = -4$. So $x^2-4x-12 = (x+2)(x-6)$.

Step 3: trinomials $ax^2 + bx + c$ (the ac-method)

For leading coefficient $a \ne 1$: find two integers $p$ and $q$ with $pq = ac$ and $p+q = b$. Then split the middle term and factor by grouping.

Example: $2x^2 + 7x + 3$. Here $a=2$, $b=7$, $c=3$, so $ac = 6$. Need $pq=6$, $p+q=7$: the pair $(1,6)$ works.

$2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (x+3)(2x+1).$

Check: $(x+3)(2x+1) = 2x^2 + x + 6x + 3 = 2x^2 + 7x + 3$. Confirmed.

Step 4: difference of squares and perfect-square trinomials

From the Polynomials lesson, you know:

• $(a+b)(a-b) = a^2 - b^2$, so $a^2 - b^2 = (a+b)(a-b)$.
• $(a+b)^2 = a^2+2ab+b^2$, so perfect-square trinomials factor as $(a+b)^2$.
• $(a-b)^2 = a^2-2ab+b^2$.

Examples:

• $x^2 - 25 = (x+5)(x-5)$.
• $4x^2 - 9 = (2x+3)(2x-3)$.
• $x^2 + 6x + 9 = (x+3)^2$.
• $9x^2 - 12x + 4 = (3x-2)^2$.

Solving equations by factoring

Once you factor, use the zero-product property:

$x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies x = 2 \text{ or } x = 3.$

Always check both solutions. $(2)^2-5(2)+6 = 4-10+6 = 0$ and $(3)^2-5(3)+6 = 9-15+6 = 0$. Both confirmed.

What doesn't factor?

Not everything factors over the rationals. $x^2 + 1$ has no real zeros (since $x^2 \ge 0$, so $x^2 + 1 \ge 1 > 0$ always). It is irreducible over $\mathbb{Q}$ — a polynomial with no rational factorization. This is the teaser for complex numbers, which live past the edge of this stratum.