4.2 Solving by Factoring
Introduction
A surprising number of real-world applications involve equations where \(x^2\) is the highest power. These quadratic equations help with everyting from minimizing costs to finding the height of a rocket or the distance traveled by a baseball and even creating smooth computer animations.
We've solved simple quadratic equations before by taking a square root. While square roots will still play a role, when equations that have both linear \(x\) and quadratic \(x^2\) terms factoring will be one of our first tools.
Solving with the Zero-Product Property
The reason factoring is useful as a solving tool has to do with an important fact that you might be able to figure out own your own as you think about the following question:
Suppose we have two quantities, \(A\) and \(B\), and that their product is \(A \cdot B=0\). What does that tell us about \(A\) and \(B\)?
If \(A\cdot B=0\), then it must be that \(A=0\) or \(B=0\) or, perhaps, both are equal to zero. There's no alternative; one (or both) must be equal to zero, guaranteed. This is called the zero-product property.
If \(A \cdot B = 0\), then \(A = 0\) or \(B = 0\).
It's important to point out that this only works for the number \(0\), hence the name zero-product property. For instance, if \(A\cdot B=6\) then there's no guarantee that either \(A=6\) or \(B=6\), it could easily just as easily be that \(A=2\) and \(B=3\).
But when the product equals zero we can, with confidence, say that at least one of the factors must be zero. Setting each factor in a quadratic equation equal to zero will create two simple linear equations. Solving each of those will give us the solutions to the original problem.
Let's walk through a few examples.
Solve \((x+5)(x-7)=0\)
Solution
Since the product of the two factors is zero, we can use the zero-product property to find the solution by setting each equal to zero and solving.
\begin{align*} x+5 &= 0 & \text{ or } & & x-7 &= 0 \\ x &= -5 & \text{ or } & & x &= 7 \end{align*}The two solutions are \(x=-5\) and \(x=7\).
Solve \(2x(3x-4)=0\)
Solution
Using the zero-product property, we set each factor equal to zero:
\begin{align*} 2x &= 0 & \text{ or } & & 3x-4 &= 0 \\ x &= 0 & \text{ or } & & x &= \frac{4}{3} \end{align*}The two solutions are \(x=0\) and \(x=\frac{4}{3}\).
This process of solving factored equations by the zero-product property can be extended naturally to equations with more than two factors. We simply set each factor, no matter how many we have, equal to zero and then solve, as we'll see in the next example.
Solve \((x+2)(x-6)(x-9)=0\)
Solution
Even though we have three factors, we still set each factor equal to zero and solve:
\begin{align*} x+2 &= 0 & \text{ or } & & x-6 &= 0 & \text{ or } & & x-9 &= 0 \\ x &= -2 & \text{ or } & & x &= 6 & \text{ or } & & x &= 9 \end{align*}The three solutions are \(x=-2\), \(x=6\), and \(x=9\).
Solving a Quadratic Equation by Factoring
In the prior examples, the equations were already factored. Most of the time, however, we will have to find the factored form ourselves.
As a refresher, recall that factoring entails finding expressions that can be multiplied together to give the original expression. We learned about two types of factoring in section 2.6 and a given problem could require one or a combination of both.
The first type was factoring out the greatest common factor (GCF). This should be used when all terms share a common factor. For instance, in \(3x^2 + 6x\), the GCF is \(3x\) allowing us to factor it as \(3x(x+2)\).
The second method deals with the special case of a trinomial in the form \(x^2+bx+c\). In such cases, we try to find two numbers that multiply to \(b\) and add up to \(c\). Take \(x^2+5x+6\) as an example. It factors as \((x+2)(x+3)\) since \(2\cdot 3=6\) and \(2+3\)=5.
If we can factor an expression that's equal to zero, then we can finish solving it using the zero-product priciple as we did earlier.
Solve the equation \(x^2+x-6=0\) by factoring.
Solution
To factor, we need two numbers that multiply to \(-6\) and add up to \(1\). These numbers are \(3\) and \(-2\).
\begin{align*} x^2 + x - 6 &= 0 \\ (x+3)(x-2) &= 0 \end{align*}By the zero-product property:
\begin{align*} x+3 &= 0 & \text{ or } & & x-2 &= 0 \\ x &= -3 & \text{ or } & & x &= 2 \end{align*}The solutions are \(x=-3\) and \(x=2\).
This next example can actually be solved two ways. Let's solve it first by factoring.
Solve \(x^2-100=0\).
Solution
The left side can be factored as a difference of squares.
\begin{align*} x^2 - 100 &= 0 \\ (x+10)(x-10) &= 0 \end{align*}By the zero-product property:
\begin{align*} x+10 &= 0 & \text{ or } & & x-10 &= 0 \\ x &= -10 & \text{ or } & & x &= 10 \end{align*}The solutions are \(x=-10\) and \(x=10\).
This example could also have been solved using square roots if we had first rewritten it at \(x^2=100\).
It's not uncommon to be able to solve a problem in multiple ways. Mathematicians are always looking for new, elegant methods to solve problems, even old ones that have been solved before. The Pythagorean Theorem, for instance, which is likely almost 4000 years old, is known to have over 300 different proofs. So feel free to explore different techniques and ideas when solving equations.
Throughout math, whenever we encounter a new problem we always want to turn it into a problem we've solved before.
In the next example the equation is not equal to zero, so what do we do? We move terms to the other side of the equation so that it does equal zero. After doing that it will look similar to problems we've solved earlier and we'll proceed as we did before.
Solve \(4x^2=28x\).
Solution
We'll have to set the equation equal to \(0\) by subtracting \(28x\) from both sides of the equation.
\begin{align*} 4x^2 &= 28x \\ 4x^2 - 28x &= 0 \\ 4x(x - 7) &= 0 \end{align*}By the zero-product property:
\begin{align*} 4x &= 0 & \text{ or } & & x-7 &= 0 \\ x &= 0 & \text{ or } & & x &= 7 \end{align*}The solutions are \(x=0\) and \(x=7\).
In this last example, it might appear that the equation is beyond our skills. But if we work with it carefully, we can turn it into something that looks more familiar.
Solve \(5x^3+10x^2-75x = 0\).
Solution
\begin{align*} 5x^3+10x^2-75x &= 0 \\ 5x(x^2 + 2x -15) &= 0 \\ 5x(x+5)(x-3) &= 0 \end{align*}
By the zero-product property:
\begin{align*} 5x &= 0 & \text{ or } & & x+5 &= 0 & \text{ or } & & x-3 &= 0 \\ x &= 0 & \text{ or } & & x &= -5 & \text{ or } & & x &= 3 \end{align*}The solutions are \(x=0\), \(x=-5\) and \(x=3\).
We now have a process for trying to solve an equation by factoring. Here's a summary of the steps:
- Set Up: Use addition and/or subtraction to move all terms to one side of the equation so the equation is equal to zero.
- Factor: Factor the expression.
- Zero-Product Property: Set each factor equal to zero separately.
- Solve: Solve for the variable in each equation from step 3.
Looking Ahead
Factoring is a powerful tool for solving quadratic equations, but it has one significant limitation: not all quadratic expressions can be factored using integers. When factoring fails or becomes too difficult to spot, we need a method that works for every quadratic equation. That final tool, the quadratic formula, is our next section.