Solving Quadratics by Factoring

Factoring is a change of form

The last lesson introduced a quadratic function as one object wearing three 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. The third form,

is factored form, and the question it answers for free is the one this lesson lives on: where is $f$ equal to zero? 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

Here is the whole engine of this lesson, stated for two real numbers.

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. The reason is that on the clock $3$ has no reciprocal: 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.

Now run the property on an equation. 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

and one quadratic equation has become two linear equations, which Chapter 1 taught you to solve. In the language of Chapter 3, where a solution set became an object in its own right, the quadratic’s solution set is the union of the two linear solution sets: $\{2\} \cup \{3\} = \{2, 3\}$. Note the contrast with a system of equations: a system joins its equations with “and,” so its solution set is an intersection. The factors here are joined by “or,” so the sets unite.

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.

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 is promiscuous about its factorizations. 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 $6$ tells you nothing at all about either number individually. Now suppose the right-hand side is not $0$. Then the step “set each factor equal to the right-hand side” is not a small abuse of the method; it is unjustified from the first symbol.

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 is the Zero Product Property’s evil twin: instead of using a zero factor, it destroys one. 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 that they coincide. A zero of the function $f$ is an input where $f(x) = 0$; 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. An $x$-intercept of the parabola $y = f(x)$ is a point where the graph meets the $x$-axis. A point sits on the $x$-axis exactly when its $y$-coordinate is $0$; for a graph point $(x, f(x))$, that says $f(x) = 0$. Three names, one collection of numbers: the $x$-coordinates of the crossings.

The figure shows one more coincidence that is not a coincidence. The last lesson derived the axis of symmetry $x = -\tfrac{b}{2a}$ from standard form. Expand the factored form and compare coefficients:

so $b = -a(r_1 + r_2)$ and $c = a\,r_1 r_2$. Then

the midpoint of the two zeros. The axis of symmetry runs exactly halfway between the crossings. You can also see that from the mirror symmetry of the parabola: the two crossings are reflections of each other, so the mirror stands midway. (The comparison $b = -a(r_1 + r_2)$, $c = a\,r_1 r_2$ is a two-way street between roots and coefficients, and the lesson on sums and products of roots will make a systematic tool of it. Here we only needed the midpoint.)

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, so read the signs off the example $f(x) = (x + 1)(x - 3)$. The two factors are both negative left of $-1$ (product positive), have opposite signs between $-1$ and $3$ (product negative), and are both positive right of $3$ (product positive). A quadratic can switch sign only by passing through zero, so the crossings carve the axis into intervals of constant sign. That is exactly what the figure shows: the arch dips below the axis precisely between its two crossings. The lesson on quadratic inequalities is built on reading signs this way.

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 two statements separate in your head, because they are about different number systems, and mixing them produces false sentences. A quadratic can have real zeros and still refuse to factor over the rationals; the next section exhibits one. 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

The hunt is finite, since only finitely many integer pairs multiply to $c$, and that finiteness is why it feels mechanical. 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.

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 crosses 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: it comes up empty whenever the zeros are irrational, and also, of course, when there are no real zeros at all. And a failed search cannot tell you which of those 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 it rewrites any quadratic in vertex form. Run once in general symbols, it produces the quadratic formula, which is why that formula is a theorem rather than a recipe. The discriminant inside that formula is the single number that announces in advance which story the graph tells: two crossings, one touch, or none. And in Chapter 5 the number system itself is enlarged, so that even $x^2 + x + 1$ surrenders a factorization there.

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. At a simple zero the sign of $f$ flips as you pass the crossing; 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 minimum value $0$ occurs at $x = r$, 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