Factoring Polynomials

Main idea

Factoring is multiplying run backwards: it rewrites a polynomial as a product. Products are what you want, because a product is zero only when one of its factors is, and common factors can cancel.

{\htmlData{sym=0}{a^2 - b^2}} = (a - b)(a + b), \qquad a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

What the symbols mean

$a^2 - b^2$
Difference of squares
Factors; a sum of squares doesn't (over the reals).

Explanation

Factoring is simply asking "what got multiplied together to make this?" It's like knowing $12 = 3\cdot4$ , but for polynomials; the pieces you multiply are called factors, hence the name. It's the main tool for solving polynomial equations and simplifying fractions. The formulas up top are just patterns you'll recognize: $a^2 - b^2$ (one square minus another) always breaks into $(a - b)(a + b)$ (the difference times the sum), and the cube formulas are the same idea one power up. It looks like a lot of rules, but it's a short checklist.

Why bother? Last lesson you multiplied $(x - 2)(x - 3)$ out to get $x^2 - 5x + 6$ . Factoring goes the other way. The expanded form is easy to evaluate, but the factored form tells you things: $(x - 2)(x - 3) = 0$ forces $x = 2$ or $x = 3$ , since a product can only be zero if one of its pieces is.

Where the trinomial trick comes from. Multiply out $(x + p)(x + q)$ and you get $x^2 + (p + q)x + pq$ . So to factor $x^2 + bx + c$ , you need two numbers $p$ and $q$ that multiply to $c$ and add to $b$ . That's not a trick to memorize; it's just reading the expansion backwards.

A checklist:

Greatest common factor first: $6x^3 - 9x^2 = 3x^2(2x - 3)$ . It makes everything after it smaller.
Special forms: difference of squares, perfect squares, sums and differences of cubes. These are the special products from last lesson, read backwards.
Trinomials $x^2 + bx + c$ : find two numbers that multiply to $c$ and add to $b$ . For $ax^2 + bx + c$ , find numbers multiplying to $ac$ , split the middle term, and group.
Grouping for four terms.

You can always check by multiplying back out, so there's never a reason to hand in a wrong factorization.

See it

This is the parabola x² − 5x + c as you slide c, and it shows that factoring and solving are the same job: it crosses the x-axis at r₁ and r₂ exactly when it factors as (x − r₁)(x − r₂). At c = 6 the crossings are 2 and 3, the factors you'd find by hand. At c = 6.25 they merge into one double root, (x − 2.5)².

Worked examples

Example 1

Factor $6x^2 + x - 2$ .

1
Here $ac = 6\cdot(-2) = -12$ , and we need two numbers that multiply to $-12$ and add to the middle coefficient $1$ . That's $4$ and $-3$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Factor $x^2 - 5x + 6$ . Give its roots.

Problem 2+5 XP

Factor $6x^3 - 9x^2$ (enter the factored form).

Problem 3+5 XP

Factor $4x^2 - 25$ (enter the factored form).

Problem 4+5 XP

Factor $6x^2 + x - 2$ (enter the factored form).

Problem 5+5 XP

Factor $x^3 - 8$ (enter the factored form).

Problem 6+5 XP

An open box is made from a 12 cm by 12 cm sheet by cutting $x$ cm squares from the corners, so its volume is $4x^3 - 48x^2 + 144x$ . Factor that completely.

Problem 7+5 XP

Solve $x^3 - x^2 - 6x = 0$ by factoring.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Factor $x^4 + 4$ into two quadratics with integer coefficients.

Where it's used

Factoring finds where a profit is zero (break-even points), when a projectile lands, and it's the first step of partial fractions in calculus.

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