2.1 — Solving by Factoring

The Question Flips

In Chapter 1, we started with data or a rule and asked: what does the curve look like? Now the question reverses. Someone hands you a quadratic equation like

$$x^2 - 5x + 6 = 0$$

and asks: for which values of $x$ is this true?

Geometrically, this is asking: where does the parabola $y = x^2 - 5x + 6$ cross the $x$-axis? Those crossing points — the roots or solutions — are where the output is exactly zero.

Finding roots is one of the most important tasks in all of mathematics. And the first method is surprisingly simple — if you notice the right structure.

A Puzzle About Zero

Here's a small but powerful fact:

If two numbers multiply to zero, then at least one of them must be zero.

This sounds obvious, but pause and appreciate it. It doesn't work for any other number. If $a \cdot b = 6$, you can't say much — maybe $a = 2$ and $b = 3$, or $a = 1$ and $b = 6$, or $a = -3$ and $b = -2$. There are infinitely many possibilities.

But if $a \cdot b = 0$? Then either $a = 0$, or $b = 0$ (or both). No wiggle room.

This is called the zero-product property, and it turns multiplication into a tool for solving equations — if we can write our quadratic as a product.

Factoring in Action

Look at $x^2 - 5x + 6$ again. Can we write it as a product of two simpler expressions?

We need two numbers that: - multiply to $6$ (the constant term) - add to $-5$ (the coefficient of $x$)

Before reading on, try to find them. Which two numbers multiply to $6$ and add to $-5$?

Check your answer

$-2$ and $-3$. They multiply to $(-2)(-3) = 6$, and they add to $-2 + (-3) = -5$.

So we can write:

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

Now the original equation becomes:

$$(x - 2)(x - 3) = 0$$

Apply the zero-product property: either $x - 2 = 0$ or $x - 3 = 0$. So:

$$x = 2 \quad \text{or} \quad x = 3$$

That's it. Two roots. The parabola $y = x^2 - 5x + 6$ crosses the $x$-axis at $x = 2$ and $x = 3$.

Verify it yourself. Plug $x = 2$ into $x^2 - 5x + 6$: you get $4 - 10 + 6 = 0$. Plug in $x = 3$: you get $9 - 15 + 6 = 0$. Both check out.

The Method, Step by Step

Here's the general procedure for solving $ax^2 + bx + c = 0$ by factoring (when $a = 1$):

1. Find two numbers $p$ and $q$ that multiply to $c$ and add to $b$.
2. Write the quadratic as $(x + p)(x + q) = 0$.
3. Apply the zero-product property: $x = -p$ or $x = -q$.

When $a = 1$, the whole problem reduces to finding two numbers with a given product and sum. That's it.

Check your understanding

Factor and solve: $x^2 + 7x + 12 = 0$

What two numbers multiply to $12$ and add to $7$?

Check your answer

$3$ and $4$. They multiply to $12$ and add to $7$. $$x^2 + 7x + 12 = (x + 3)(x + 4) = 0$$ So $x = -3$ or $x = -4$. Verify: $(-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0$. $(-4)^2 + 7(-4) + 12 = 16 - 28 + 12 = 0$. Both work.

Recognizing Common Patterns

Not every factoring problem requires searching for two mystery numbers. Some quadratics wear their structure on the surface. Learning to recognize these patterns speeds things up enormously.

Difference of Squares

Look at this equation:

$$x^2 - 9 = 0$$

There's no middle term — no $bx$. And $9 = 3^2$. This is a difference of squares:

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

So $x^2 - 9 = (x + 3)(x - 3) = 0$, giving $x = 3$ or $x = -3$.

Why does this work? Multiply out $(x + 3)(x - 3)$: you get $x^2 - 3x + 3x - 9 = x^2 - 9$. The middle terms cancel — that's the signature of a difference of squares. Whenever the $b$ coefficient is zero and $c$ is a negative perfect square, you're looking at this pattern.

Here's the general form: whenever you see $x^2 - k^2$, you can immediately write $(x + k)(x - k) = 0$ and read off $x = k$ or $x = -k$.

Perfect Square Trinomials

Now look at this one:

$$x^2 - 6x + 9 = 0$$

The two numbers that multiply to $9$ and add to $-6$ are... $-3$ and $-3$. The same number, twice. So:

$$x^2 - 6x + 9 = (x - 3)(x - 3) = (x - 3)^2 = 0$$

This gives just one solution: $x = 3$ (a repeated root). Geometrically, the parabola doesn't cross the $x$-axis — it just touches it at a single point and bounces back.

The pattern: $x^2 - 2kx + k^2 = (x - k)^2$. The telltale sign is that the constant term is the square of half the middle coefficient. Here: half of $-6$ is $-3$, and $(-3)^2 = 9$. That's a match.

Quick check: Is $x^2 + 10x + 25$ a perfect square trinomial? Half of $10$ is $5$, and $5^2 = 25$. Yes — it's $(x + 5)^2$.

When Factoring Doesn't Work

Try to factor $x^2 + 3x + 1 = 0$.

You need two numbers that multiply to $1$ and add to $3$. The only pairs that multiply to $1$ are $(1, 1)$ and $(-1, -1)$. Their sums are $2$ and $-2$. Neither is $3$.

This equation has solutions — they just aren't integers. Factoring (with whole numbers) won't find them.

An honest limitation. Factoring is fast and elegant when it works, but it only works when the roots are "nice" numbers. Many real-world quadratics don't have integer roots. In the next two sections, we'll develop methods — completing the square and the quadratic formula — that work for every quadratic, nice or not.

Recognizing when factoring applies and when it doesn't is itself an important skill. If you spend more than a minute hunting for the right pair of numbers and can't find them, that's the equation telling you to use a different tool.

Reflection

Let's step back and name the key idea.

Solving a quadratic by factoring rests on one principle: zero is special in multiplication. Unlike any other number, zero "infects" a product — if any factor is zero, the whole product is zero. We exploit this by rewriting $ax^2 + bx + c$ as a product of two linear factors, then setting each one to zero.

The method is fast but limited. It works beautifully when the roots are integers (or simple fractions), and it fails quietly when they aren't. The patterns — difference of squares, perfect square trinomials — are shortcuts for recognizing structure that's already there.

Connection to what you know: Factoring is really the reverse of multiplication. In Chapter 1, you multiplied expressions like $(x + 2)(x + 3)$ to get $x^2 + 5x + 6$. Factoring starts with $x^2 + 5x + 6$ and recovers $(x + 2)(x + 3)$. Every time you factor, you're un-doing a multiplication.

Practice

Problem 1 — Concrete. Solve by factoring:

(a) $x^2 + 5x + 6 = 0$

(b) $x^2 - 7x + 10 = 0$

(c) $x^2 - 4x - 12 = 0$

Solutions

(a) Need two numbers that multiply to $6$ and add to $5$: that's $2$ and $3$. $$(x + 2)(x + 3) = 0 \implies x = -2 \text{ or } x = -3$$ (b) Multiply to $10$, add to $-7$: that's $-2$ and $-5$. $$(x - 2)(x - 5) = 0 \implies x = 2 \text{ or } x = 5$$ (c) Multiply to $-12$, add to $-4$: that's $-6$ and $2$. $$(x - 6)(x + 2) = 0 \implies x = 6 \text{ or } x = -2$$

Problem 2 — Patterns. Solve by recognizing the pattern (don't hunt for pairs — use the structure):

(a) $x^2 - 25 = 0$

(b) $x^2 + 8x + 16 = 0$

(c) $4x^2 - 1 = 0$

Solutions

(a) Difference of squares: $(x + 5)(x - 5) = 0$. Solutions: $x = 5$ or $x = -5$. (b) Perfect square trinomial: half of $8$ is $4$, and $4^2 = 16$. So $(x + 4)^2 = 0$, giving $x = -4$ (repeated root). (c) Difference of squares again, but with $a \neq 1$: $4x^2 - 1 = (2x)^2 - 1^2 = (2x + 1)(2x - 1) = 0$. Solutions: $x = \frac{1}{2}$ or $x = -\frac{1}{2}$.

Problem 3 — Debug. A student solves $x^2 = 3x$ by dividing both sides by $x$, getting $x = 3$. They submit this as the only solution. What went wrong?

Solution

Dividing both sides by $x$ silently assumes $x \neq 0$ — and throws away the solution $x = 0$. The correct approach: move everything to one side first. $$x^2 - 3x = 0 \implies x(x - 3) = 0 \implies x = 0 \text{ or } x = 3$$ **Lesson:** Never divide both sides of an equation by a variable — you might be dividing by zero and losing a solution. Instead, factor.

Problem 4 — Decision. For each equation, decide whether factoring (with integers) will work. If it will, solve. If not, explain how you know.

(a) $x^2 - 2x - 15 = 0$

(b) $x^2 + x + 1 = 0$

(c) $x^2 - 1 = 0$

Solutions

(a) Need two numbers that multiply to $-15$ and add to $-2$: that's $-5$ and $3$. Factoring works. $$(x - 5)(x + 3) = 0 \implies x = 5 \text{ or } x = -3$$ (b) Need two numbers that multiply to $1$ and add to $1$. The only pairs with product $1$ are $(1, 1)$ and $(-1, -1)$, with sums $2$ and $-2$. Neither equals $1$. **Factoring doesn't work.** (This equation actually has no real solutions at all — its discriminant is $1 - 4 = -3 < 0$.) (c) Difference of squares: $(x + 1)(x - 1) = 0$. Solutions: $x = 1$ or $x = -1$.

What's Next

Factoring is fast when the roots are nice, but most quadratics in the real world don't have integer roots. We need a method that always works. In the next section, we'll discover completing the square — a geometric idea that transforms any quadratic into a form we can solve by inspection.