Table of Contents
Understanding the Factoring Method for Quadratic Equations
In this chapter, we focus on solving quadratic equations by factoring. The general idea of a quadratic equation and other solution methods are handled elsewhere; here we concentrate on when and how factoring works.
A quadratic equation in standard form looks like
$$
ax^2 + bx + c = 0,
$$
where $a$, $b$, and $c$ are real numbers and $a \ne 0$.
The factoring method tries to rewrite the left side as a product of simpler expressions. Once it is written as a product, we use the zero product property:
If $PQ = 0$, then $P = 0$ or $Q = 0$ (or both).
So the main steps are:
- Write the quadratic in standard form $ax^2 + bx + c = 0$.
- Factor the quadratic expression on the left.
- Set each factor equal to zero and solve the resulting simple equations.
All of the techniques below are about step 2: how to factor.
1. Recognizing When Factoring Is Useful
The factoring method works well when the quadratic expression $ax^2 + bx + c$ can be written as a product of simple factors with “nice” numbers (usually integers or simple fractions).
For example:
- $x^2 - 5x + 6 = 0$ factors nicely and is easy to solve.
- $x^2 - 2x + 2 = 0$ does not factor nicely over the integers; other methods are usually better for such equations.
In practice, you often try factoring first. If you cannot find factors after a reasonable attempt, you switch to another method (such as the quadratic formula).
2. Factoring Quadratics with $a = 1$
These are equations of the form
$$
x^2 + bx + c = 0.
$$
To factor $x^2 + bx + c$, you look for two numbers that:
- multiply to $c$ (the constant term), and
- add to $b$ (the coefficient of $x$).
Then you can write:
$$
x^2 + bx + c = (x + m)(x + n),
$$
where $m$ and $n$ are those two numbers.
Example: Positive $c$
Solve
$$
x^2 - 5x + 6 = 0.
$$
- Find two numbers that multiply to $6$ and add to $-5$.
- Possible factor pairs of $6$: $(1,6)$ and $(2,3)$.
- To get a sum of $-5$, both must be negative: $(-2, -3)$.
- Check: $(-2)\cdot(-3) = 6$ and $-2 + (-3) = -5$.
- Factor:
$$
x^2 - 5x + 6 = (x - 2)(x - 3).
$$ - Set each factor equal to zero:
$$
x - 2 = 0 \quad \text{or} \quad x - 3 = 0.
$$
So
$$
x = 2 \quad \text{or} \quad x = 3.
$$
Example: Negative $c$
Solve
$$
x^2 + x - 6 = 0.
$$
- Look for two numbers that multiply to $-6$ and add to $1$.
- Factor pairs of $-6$: $(1,-6)$, $(-1,6)$, $(2,-3)$, $(-2,3)$.
- $-2$ and $3$ work: $(-2)\cdot3 = -6$, and $-2 + 3 = 1$.
- Factor:
$$
x^2 + x - 6 = (x - 2)(x + 3).
$$ - Solve:
$$
x - 2 = 0 \Rightarrow x = 2,\quad x + 3 = 0 \Rightarrow x = -3.
$$
3. Using a Greatest Common Factor (GCF) First
Before factoring a quadratic completely, always look for a common factor in all terms. This makes the quadratic simpler.
Example
Solve
$$
6x^2 - 12x = 0.
$$
- Factor out the GCF (which is $6x$):
$$
6x^2 - 12x = 6x(x - 2).
$$ - Now the equation is
$$
6x(x - 2) = 0.
$$ - Use the zero product property:
$$
6x = 0 \quad \text{or} \quad x - 2 = 0.
$$
So
$$
x = 0 \quad \text{or} \quad x = 2.
$$
Sometimes this first step of pulling out the GCF is the only factoring needed. Other times, you pull out the GCF and then factor the remaining quadratic further.
4. Factoring Quadratics with $a \ne 1$ (Integer Coefficients)
Now consider quadratics like
$$
ax^2 + bx + c,
$$
where $a \ne 1$ and $a, b, c$ are integers.
You want to write
$$
ax^2 + bx + c = (px + q)(rx + s)
$$
with integers $p, q, r, s$, and typically $pr = a$ and $qs = c$.
There are two common approaches:
- Trial and error (guess and check) with factor pairs.
- The “product–sum” (or $ac$) method.
We focus on the product–sum method because it is more systematic.
The Product–Sum ($ac$) Method
Given $ax^2 + bx + c$:
- Compute $ac$ (multiply the first and last coefficients).
- Find two integers $m$ and $n$ such that:
- $m \cdot n = ac$,
- $m + n = b$.
- Rewrite the middle term $bx$ as $mx + nx$.
- Factor by grouping.
Example
Solve
$$
2x^2 + 7x + 3 = 0.
$$
- Compute $ac$: $a = 2$, $c = 3$, so $ac = 6$.
- Find $m$ and $n$ so that $m n = 6$ and $m + n = 7$.
- $1 \cdot 6 = 6$ and $1 + 6 = 7$.
- Rewrite $7x$ as $x + 6x$:
$$
2x^2 + 7x + 3 = 2x^2 + x + 6x + 3.
$$ - Group and factor each pair:
$$
(2x^2 + x) + (6x + 3).
$$
Factor:
$$
2x^2 + x = x(2x + 1),\quad 6x + 3 = 3(2x + 1).
$$
So
$$
2x^2 + 7x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).
$$ - Set each factor equal to zero:
$$
x + 3 = 0 \Rightarrow x = -3, \quad 2x + 1 = 0 \Rightarrow x = -\tfrac12.
$$
Another Example (with a Negative $c$)
Solve
$$
3x^2 - 2x - 8 = 0.
$$
- $a = 3$, $c = -8$, so $ac = -24$.
- Find $m, n$ with $mn = -24$, $m + n = -2$.
- $4$ and $-6$ work: $4 \cdot (-6) = -24$, $4 + (-6) = -2$.
- Rewrite $-2x$ as $4x - 6x$:
$$
3x^2 - 2x - 8 = 3x^2 + 4x - 6x - 8.
$$ - Group:
$$
(3x^2 + 4x) + (-6x - 8).
$$
Factor:
$$
3x^2 + 4x = x(3x + 4),\quad -6x - 8 = -2(3x + 4).
$$
Combine:
$$
3x^2 - 2x - 8 = x(3x + 4) - 2(3x + 4) = (x - 2)(3x + 4).
$$ - Solve:
$$
x - 2 = 0 \Rightarrow x = 2,\quad 3x + 4 = 0 \Rightarrow x = -\tfrac43.
$$
5. Factoring Special Quadratic Forms
Some quadratic equations fit into special patterns that factor quickly. Recognizing these patterns saves time.
5.1 Perfect Square Trinomials
These are expressions like:
- $x^2 + 2ax + a^2 = (x + a)^2$,
- $x^2 - 2ax + a^2 = (x - a)^2$.
Example
Solve
$$
x^2 + 6x + 9 = 0.
$$
Recognize $6x = 2 \cdot 3 \cdot x$ and $9 = 3^2$, so this matches
$$
x^2 + 2\cdot 3 x + 3^2 = (x + 3)^2.
$$
So
$$
(x + 3)^2 = 0.
$$
Then
$$
x + 3 = 0 \Rightarrow x = -3.
$$
There is a repeated root here: both solutions are the same.
5.2 Difference of Squares
A very common pattern is:
$$
A^2 - B^2 = (A - B)(A + B).
$$
If a quadratic equation can be written as a difference of two squares, it factors immediately.
Example
Solve
$$
x^2 - 16 = 0.
$$
Here $x^2 - 16 = x^2 - 4^2$, so
$$
x^2 - 16 = (x - 4)(x + 4).
$$
Solve:
$$
x - 4 = 0 \Rightarrow x = 4,\quad x + 4 = 0 \Rightarrow x = -4.
$$
Another Example (with a coefficient)
Solve
$$
4x^2 - 25 = 0.
$$
Write $4x^2 = (2x)^2$ and $25 = 5^2$, so
$$
4x^2 - 25 = (2x)^2 - 5^2 = (2x - 5)(2x + 5).
$$
Then
$$
2x - 5 = 0 \Rightarrow x = \tfrac52,\quad 2x + 5 = 0 \Rightarrow x = -\tfrac52.
$$
6. Factorable vs. Non-Factorable (Over the Integers)
Not every quadratic expression with integer coefficients can be factored nicely into linear factors with integer coefficients.
For example:
- $x^2 - x - 2$ factors as $(x - 2)(x + 1)$.
- $x^2 - 2x + 2$ does not factor into $(x + m)(x + n)$ with integers $m$, $n$.
In practice, when you try to factor and no combination works:
- First, check if a GCF can be factored out.
- Second, try the product–sum method. If you cannot find integers $m, n$ with $mn = ac$ and $m + n = b$, the quadratic is not factorable over the integers.
- In such a case, you solve it using another method (for example, the quadratic formula), not by factoring.
7. Solving Quadratic Equations: Using the Factors
After factoring the left side, the actual solving step is straightforward:
- Ensure the equation is written as “something equals zero,” for example
$$
(2x - 3)(x + 5) = 0.
$$ - Apply the zero product property:
$$
2x - 3 = 0 \quad \text{or} \quad x + 5 = 0.
$$ - Solve each linear equation:
$$
2x - 3 = 0 \Rightarrow x = \tfrac32,\quad x + 5 = 0 \Rightarrow x = -5.
$$
You can always check your solutions by substituting them back into the original quadratic equation and verifying that they make the equation true.
8. Typical Pitfalls to Avoid
- Forgetting to set the equation equal to zero before factoring. You should always rearrange to the form $ax^2 + bx + c = 0$.
- Dropping a factor when using the zero product property. Every factor set to zero gives a possible solution.
- Not factoring out the GCF first, which can hide simpler factoring patterns.
- Sign mistakes when searching for two numbers that multiply to $ac$ and add to $b$. Carefully check multiplication and addition.
- Assuming everything factors nicely. If no suitable factors exist, stop and use another method instead of forcing incorrect factors.
9. Summary of the Factoring Method
To solve a quadratic equation by factoring:
- Rewrite in standard form:
$$
ax^2 + bx + c = 0.
$$ - Factor completely:
- First take out any GCF.
- Use:
- simple product–sum method when $a = 1$,
- $ac$ method or trial and error when $a \ne 1$,
- special patterns like perfect square trinomials and difference of squares when applicable.
- Set each factor equal to zero and solve the resulting linear equations.
- Check your solutions in the original equation.
This factoring method is especially powerful when the quadratic has integer solutions and its coefficients are not too complicated. It is often the fastest way to solve such equations.