Lesson · Quadratics and Complex Numbers · Solving quadratics

Solving by Factoring

With a zero on one side and a factorisation on the other, the zero product property finishes the job in a line. All the work is in the factoring, which is Polynomials and Factoring's, and in the two things the finished answer can hide: a constant factor that contributes nothing, and a repeated factor that contributes one solution rather than two.

Open the interactive lesson Practice

x² + 3x - 10 = 0 standard form first (x + 5)(x - 2) = 0 factor x + 5 = 0 or x - 2 = 0 zero product property x = -5 or x = 2 two solutions, both substituted back

The method, in four lines

Core ideas

What this lesson establishes

Nothing here is new. This is Polynomials and Factoring's factoring followed by “Quadratic Equations and the Zero Product Property”'s property, plus one decision: how long to keep trying.

Standard form comes first, every time

3x² = 5x - 2 has to become 3x² - 5x + 2 = 0 before anything is factored. Factoring one side of an equation whose other side is not zero achieves nothing at all, however neat the result looks.

A repeated factor gives one solution

(2x - 3)² = 0 produces the equation 2x - 3 = 0 twice, so it produces x = 3/2 once. Calling that a double root is a counting convention that keeps later statements tidy; as a solution of the equation it is one number.

Factoring is a method, not the method

It is the fastest route when the coefficients cooperate and useless when they do not. x² - 4x + 1 = 0 has two perfectly ordinary solutions that no integer factorisation will ever produce. “Completing the Square” and “The Quadratic Formula” exist for those.

The material

From a factorisation to a solution set

The procedure, the factors that contribute nothing, and what to do when the factoring will not come.

“Quadratic Equations and the Zero Product Property” supplied the licence and Polynomials and Factoring supplied the technique. Put together: write the equation as (something)(something) = 0, then set each factor to zero.

x² + 3x = 10 x² + 3x - 10 = 0 subtract 10 from both sides (x + 5)(x - 2) = 0 two numbers, product -10 and sum 3: 5, -2 x + 5 = 0 or x - 2 = 0 x = -5 or x = 2

Take the common factor out, then ignore it

2x² - 8x = 0 factors as 2x(x - 4) = 0. Three factors are written down — 2, x and x - 4 — but the first contributes nothing, because 2 = 0 is a false statement with no solutions. The answers are x = 0 and x = 4.

A constant factor never creates a solution and never destroys one. That is worth knowing in both directions: multiplying an equation through by 3 leaves its solution set alone, and so does dividing by 3. Only multiplying or dividing by something that could be zero is dangerous, which is the warning from “Quadratic Equations and the Zero Product Property”.

Repeated factors

4x² - 12x + 9 = 0 is (2x - 3)² = 0, so the only equation to solve is 2x - 3 = 0 and the only solution is x = 3/2. Substituting: 4(9/4) - 12(3/2) + 9 = 9 - 18 + 9 = 0.

It is standard to say this equation has a double root at 3/2, or a root of multiplicity 2. That is a statement about the factorisation, not about the solution set — the equation has exactly one solution. The convention earns its keep in “The Discriminant” and in “Graphs of Polynomial Functions” in Polynomials and Factoring, where counting roots with multiplicity is what makes the statements come out clean.

Theorem · What a factorisation says about the solutions

Suppose ax² + bx + c = a(x - r)(x - s) for every value of x, with a ≠ 0. Then the solutions of ax² + bx + c = 0 are exactly r and s.

Both directions are needed and both are short. Each of r and s makes a bracket zero, so each is a solution. Any other value makes both brackets non-zero, and a is non-zero, so the product is non-zero and that value is not a solution. There is nothing else to check and nothing else to find.

When the leading coefficient is not 1

Polynomials and Factoring's AC method still applies unchanged. For 6x² + x - 12 = 0, take ac = 6 · (-12) = -72 and look for two numbers with product -72 and sum 1. They are 9 and -8.

6x² + x - 12 = 0 6x² + 9x - 8x - 12 = 0 split the middle term 3x(2x + 3) - 4(2x + 3) = 0 group in pairs (2x + 3)(3x - 4) = 0 2x + 3 = 0 or 3x - 4 = 0 x = -3/2 or x = 4/3

Both answers are fractions and both are exact. -3/2 is the solution; -1.5 is a decimal that happens to equal it, and 4/3 has no decimal that equals it at all. Exact fractions are the standard on this path for that reason.

When the factoring will not come

A complete search that fails is a result about integer factorisation and nothing more. x² - 4x + 1 = 0 has no pair of integers with product 1 and sum -4, so it does not factor over the integers. It still has two solutions, 2 + √3 and 2 - √3, which “Completing the Square” produces. "It does not factor" and "it has no solutions" are different sentences, and only the first one has been established.

Example · Running the method backwards

To build an equation whose solutions are -5 and 2, start from (x + 5)(x - 2) = 0 and expand: x² + 3x - 10 = 0. Multiplying through by any non-zero constant gives another equation with the same two solutions, so 2x² + 6x - 20 = 0 works equally well.

There is therefore no such thing as the equation with a given pair of solutions. Any two quadratics with the same roots differ by a constant multiple, which is the same observation as "a constant factor contributes nothing", read the other way round.

Interactive

Solving by factoring

The factored form, found by search

Method

Solving a quadratic by factoring

Steps 1 and 2 hold all the work; step 3 is a single line.

Put the equation in standard form

Expand the brackets, clear any fractions, and move every term to one side so the other reads 0. Arrange in descending powers, so that a, b and c can be seen.

Factor completely, common factor first

Take out the GCF, as “Factoring Out the Greatest Common Factor” in Polynomials and Factoring insists, then factor what is left. A monic trinomial needs a product-and-sum pair from “Factoring Simple Trinomials” in that course; anything else needs “The AC Method”.

Set each non-constant factor to zero and solve

Each one gives a linear equation. Skip any constant factor, which can never be zero. A repeated factor gives its solution once, not twice.

Substitute every answer into the original equation

Into the equation as given, not the rearranged one, which may already carry the error. Two substitutions cost a line each and confirm the whole chain of reasoning above them.

Factor and split

Score 0 / 0

Choose the answer you can justify.

Three equations, three shapes

One needing rearrangement, one with a leading coefficient, and one with a repeated factor.

A. x² + 3x = 10 x² + 3x - 10 = 0 subtract 10 (x + 5)(x - 2) = 0 product -10, sum 3: 5 and -2 x = -5 or x = 2 check x = -5: 25 - 15 = 10 x = 2: 4 + 6 = 10 B. 6x² + x - 12 = 0 ac = -72, 9 · (-8) = -72, 9 + (-8) = 1 6x² + 9x - 8x - 12 = 0 3x(2x + 3) - 4(2x + 3) = 0 (2x + 3)(3x - 4) = 0 x = -3/2 or x = 4/3 check x = -3/2: 6(9/4) - 3/2 - 12 = 27/2 - 3/2 - 12 = 0 x = 4/3: 6(16/9) + 4/3 - 12 = 32/3 + 4/3 - 12 = 0 C. 4x² - 12x + 9 = 0 (2x - 3)² = 0 a perfect square trinomial 2x - 3 = 0 x = 3/2 one solution, counted twice check x = 3/2: 9 - 18 + 9 = 0

In A the check was carried out on x² + 3x = 10, the equation as given. Checking against x² + 3x - 10 = 0 instead would agree with itself even if the subtraction had gone wrong, which is the one thing a check is supposed to catch.

C has one solution, and nothing was lost in getting there. (2x - 3)² gives the equation 2x - 3 = 0 twice, and solving the same equation twice returns the same number twice. “Graphs of Quadratic Functions” shows what this looks like on a graph: a curve that touches the x-axis instead of crossing it.

For a faded AC-method solve, take 12x² - 11x - 5 = 0. The supplied search target is product -60 and sum -11; use -15 and 4 to split the middle term. Group, solve both linear factors and check the two values in the original. You should reach x = 5/4 and x = -1/3; an inverted fraction identifies the factor-solving step that needs repair.

Common mistakes

Factoring before there is a zero on the other sidex² + 3x = 10 factors as x(x + 3) = 10, and that is a dead end. Standard form is not a tidiness convention here; it is the hypothesis of the theorem you are about to use.

Inverting a factor instead of solving it3x - 4 = 0 gives x = 4/3, not x = 3/4 and not x = -4/3. Solve the small equation properly: add 4, then divide by 3. Reading the numerator and denominator off the bracket is a habit that works by accident on (x - 4) and fails everywhere else.

Reporting "does not factor" as "no solutions"x² - 4x + 1 = 0 has no integer factorisation and two real solutions, 2 ± √3. The failure of a search over the integers says nothing at all about whether solutions exist.

Check your understanding

✓

Finish when factoring is a first attempt you abandon quickly.

You should reach standard form without thinking, spend a few seconds looking for an integer factorisation, and move to “Completing the Square” or “The Quadratic Formula” without regret when none appears. Every answer checked in the original equation, and a repeated factor reported as one solution.

Note: Factoring is the fastest method and the least reliable, because it works only when the coefficients were chosen to let it. “The Square Root Property”, “Completing the Square” and “The Quadratic Formula” remove the condition a piece at a time: the square root property handles anything with no x term, completing the square handles everything, and the formula is completing the square carried out once and kept.