Solving Quadratics by Factoring

Factoring is a change of form

The last lesson introduced a quadratic function as one object wearing several normal forms, each of which answers one question for free. Standard form $f(x) = ax^2 + bx + c$ hands you the $y$-intercept, since $f(0) = c$, and the end behavior, from the sign of $a$. Vertex form $f(x) = a(x - h)^2 + k$ hands you the vertex and everything that flows from it. Standard form and vertex form exist for every quadratic. A third form,

is factored form, and where it exists over the real numbers, it answers for free the question this lesson lives on: where is $f$ equal to zero? Not every quadratic has one, since not every quadratic has a real zero to display; the section “One zero buys one factor” pins down exactly when it does. Factoring is not a new kind of problem. It is the change of form that starts at standard and ends at factored, done precisely because you want the zero question answered.

Half of the claim that “the zeros are visible” costs nothing to check. Substitute $x = r_1$: the first factor becomes $r_1 - r_1 = 0$, so the whole product is $f(r_1) = a \cdot 0 \cdot (r_1 - r_2) = 0$. The same happens at $x = r_2$. So $r_1$ and $r_2$ are zeros of $f$, by substitution alone.

But “visible” claims more than that. It claims $r_1$ and $r_2$ are the only zeros, that no third number sneaks in. Substitution cannot rule out numbers you have not tried. What closes the gap is the next section, and it is a fact about the real numbers themselves, not about polynomials.

The Zero Product Property

To solve $(x - 2)(x - 3) = 0$, read the left side as a product of the two real numbers $x - 2$ and $x - 3$. The product is zero, so at least one of those two numbers must itself be zero:

and one quadratic equation has become two linear equations, which Chapter 1 taught you to solve. The two equations are joined by “or,” not “and”: either factor being zero is enough, so the solution set collects every number that solves either one, $\{2, 3\}$.

That move rested on one fact: a product of two real numbers is zero only when one of the numbers itself is zero. It is called the Zero Product Property, and it is the whole engine of this lesson. Here it is stated precisely, for two real numbers, and proved.

Notice what the proof leaned on: every nonzero real number has a reciprocal. That is a privilege of the real numbers, not a law of arithmetic in general. Build a toy arithmetic out of a clock face, where you count hours and every multiple of $12$ lands back at $0$. In that system $3 \times 4 = 12 = 0$: a product of two nonzero numbers that is zero. The property fails there, and it fails exactly where the proof would break: on the clock $3$ has no reciprocal, since multiplying $3$ by whole numbers only ever lands on $0$, $3$, $6$, or $9$, never on $1$. Number systems in which nonzero factors can multiply to zero are said to have zero divisors; the real numbers have none. Every time you set a factor equal to zero this week, that is the license you are using.

The general factored form adds nothing new. If $a \neq 0$ and $a(x - r_1)(x - r_2) = 0$, multiply by $\tfrac{1}{a}$ to get $(x - r_1)(x - r_2) = 0$, and then $x = r_1$ or $x = r_2$. So $f(x) = a(x - r_1)(x - r_2)$ vanishes exactly at $r_1$ and $r_2$: the zeros on display in factored form are all the zeros there are. The gap left by the first section is closed, and the number system is what closed it.

That checkpoint started from a factored equation. The harder, more common job is starting from standard form, the way Worked Example 1 did.

The zero on one side is not optional

The property is about zero and about nothing else. Zero is the only number that pins down a factor: the only way to multiply two reals and get $0$ is for one of them to be $0$. Every other number factors in many ways. Six, for instance, is $1 \cdot 6$, and $2 \cdot 3$, and $\tfrac{1}{2} \cdot 12$, and $(-4)\left(-\tfrac{3}{2}\right)$, and infinitely many other products. Knowing that a product of two numbers equals $6$ does not tell you what either number is. So if the right-hand side is not $0$, the step “set each factor equal to the right-hand side” is not a small shortcut; nothing justifies it.

The same discipline settles equations that arrive half-factored. For $x^2 = 5x$, dividing both sides by $x$ looks efficient and produces $x = 5$, but the division quietly assumes $x \neq 0$, and $x = 0$ happens to be a solution. Dividing by a quantity that might be zero does the opposite of what the Zero Product Property does: instead of reading a zero factor for information, it throws one away. Move everything to one side and factor instead: $x^2 - 5x = x(x - 5) = 0$, so the solution set is $\{0, 5\}$, both elements intact.

Three names for one object

Three chapters of this course have each given a name to the same numbers, and it is worth pinning down how they relate. A zero of the function $f$ is a number $r$ where $f(r) = 0$; that is Chapter 2 vocabulary. A root (or solution) of the equation $ax^2 + bx + c = 0$ is a number that makes the equation true; that is equation-solving vocabulary. Substituting shows these are the same number: a zero of $f$ is exactly a root of $f(x) = 0$. The $x$-intercept of the parabola $y = f(x)$ is different in kind: it is a point, not a number, the point where the graph meets the $x$-axis. A point sits on the $x$-axis exactly when its $y$-coordinate is $0$, so the graph point $(r, f(r))$ lands on the axis exactly when $f(r) = 0$, that is, exactly when $r$ is a zero. So every zero $r$ gives the $x$-intercept point $(r, 0)$, and every $x$-intercept point’s $x$-coordinate is a zero. Two names for one number, plus a point that number locates on the graph.

The figure shows one more coincidence that is not a coincidence. A parabola is a mirror shape: the two points where it meets the axis are reflections of each other across the axis of symmetry, so that axis has to run exactly halfway between them,

the midpoint of the two zeros. (Expanding $a(x - r_1)(x - r_2)$ and comparing it to standard form gives the same fact algebraically, and turns into a systematic tool in the lesson on sums and products of roots.)

Factored form also controls the sign of $f$ everywhere, not just where it vanishes: each factor $x - r$ is negative to the left of $r$ and positive to the right of it, and a quadratic can switch sign only by passing through zero. For $f(x) = (x + 1)(x - 3)$, both factors are negative left of $-1$ (product positive), the factors disagree between $-1$ and $3$ (product negative), and both are positive right of $3$ (product positive), which is exactly what the figure shows: the arch dips below the axis only between its two zeros. The lesson on quadratic inequalities builds on this.

One zero buys one factor

Factored form displays its zeros. This section proves the reverse: any known zero can be cashed in for a factor, so the two currencies, zeros and factors, exchange in both directions.

Keep the real and rational versions of this fact separate; they use different number systems, and mixing them up produces false statements. A quadratic can have real zeros and still refuse to factor over the rationals, as the next section shows. What can never happen is a real zero without a real factorization, or a real factorization without a real zero. The proof also hands you a working tool: find one zero by any means, even by guessing, and the factorization follows.

Factoring is a search, and the search can fail

Look honestly at what the factoring step is. To factor $x^2 + bx + c$ over the integers, you hunt for a pair of integers $p$ and $q$ with

When $c \neq 0$, the hunt is finite, since only finitely many integer pairs multiply to a fixed nonzero number, and that finiteness is why it feels mechanical. (When $c = 0$, factor out $x$ first, as the checkpoint above did.) But it is still a search, and a search can come up empty. When it does, you have learned something about your candidate list, not necessarily about the equation.

Take $x^2 - 2$. The integer pairs with product $-2$ are $1, -2$ and $2, -1$, with sums $-1$ and $1$: never $0$. So the integer hunt fails, and no rational pair rescues it either. The reason is that a rational factorization would hand us a rational zero, by the last section, and the zeros of $x^2 - 2$ are $\pm\sqrt{2}$, which are irrational. Yet over the real numbers the same polynomial factors on sight, as a difference of squares with $k = \sqrt{2}$:

So the honest sentence is never ”$x^2 - 2$ does not factor.” It is ”$x^2 - 2$ does not factor over the rationals, and does factor over the reals.” The claim has no meaning until you name the number system, because the supply of allowed coefficients is what the search draws on.

A second kind of failure cuts deeper: some quadratics do not factor even over the reals. By the previous section, that must mean those quadratics have no real zeros at all. Take $x^2 + x + 1$: add and subtract $\tfrac14$ to rewrite it as $\left(x + \tfrac12\right)^2 + \tfrac34$, a square, which is never negative, plus $\tfrac34$. So $x^2 + x + 1 \geq \tfrac34$ for every real $x$: it can never reach $0$, and the search for a real factorization was never going to succeed. It fails not for lack of cleverness, but because there is nothing to find.

The integer hunt can also fail while real zeros quietly exist. For $x^2 + 2x - 4$, no integer pair has product $-4$ and sum $2$ (check: $1, -4$ and $2, -2$ and $4, -1$ sum to $-3$, $0$, $3$). But $f(0) = -4 < 0$ while the parabola opens upward and climbs without bound on both sides, so the graph starts high, dips below the axis, and climbs high again: it meets the axis twice. Two real zeros exist, and they are simply invisible to an integer search because they are irrational.

This is the honest reason the chapter continues past this lesson. Factoring, when it works, is the fastest route to the zeros. But it is a search, and the integer or rational version of that search comes up empty whenever the zeros are irrational, or when there are no real zeros at all, and a failed search cannot tell you which of those two situations you are in. Completing the square, the next lesson, is not a search: it is a computation with no step that can fail, and the discriminant that comes out of it will announce in advance which of the three situations you are in, without a search.

Old friends, read as factored forms

Your earlier course devoted a chapter to special factorizations. In this chapter’s language, each one is a factored form you should recognize on sight, because each instantly answers the zero question.

A difference of squares is the factored form with a symmetric pair of zeros:

The symmetry is visible from standard form too: $b = 0$, so the axis of symmetry is $x = 0$, midway between $-k$ and $k$, just as the midpoint rule promised.

A perfect square trinomial is the factored form with a repeated zero:

the case the next section examines. And a common factor is a factored form that is easy to walk past: $2x^2 - 8x = 2x(x - 4)$ is already a product. The factor $x$ is just $(x - 0)$ in everyday clothes, contributing the zero $0$. That is the checkpoint above in disguise: dividing $2x^2 = 8x$ through by $x$ throws that zero away, while moving everything to one side keeps it.

A repeated root touches instead of crossing

When the two zeros coincide, the factored form collapses to

and the solution set of $f(x) = 0$ is $\{r\}$: one element. A set does not count multiplicity, so nothing about the set remembers that the factor appeared twice. The graph remembers instead.

Take $a > 0$. A square is never negative, so $a(x - r)^2 \geq 0$ for every $x$, with equality only at $x = r$. The entire parabola lies on or above the $x$-axis and meets it at the single point $(r, 0)$: the graph touches the axis and turns back without crossing. We say the parabola is tangent to the axis there. (If $a < 0$ instead, the same argument runs upside down: the parabola lies on or below the axis and is still tangent at $(r, 0)$, just from underneath.) At a simple zero the sign of $f$ flips as you pass the zero; at a repeated zero the square locks the sign, so the graph stays on one side. And the point of tangency is no ordinary point: the value $0$ at $x = r$ is the minimum (or, when $a < 0$, the maximum), so the vertex is $(r, 0)$, sitting on the axis itself. In this case factored form and vertex form are literally the same expression, $a(x - r)^2 + 0$, one more sign that the repeated root is the boundary case between two crossings and none. The discriminant, two lessons from now, will give that three-way split a single numerical name.

Common mistakes