4.2 Solving by Factoring
Introduction
A surprising number of applications in real life involve equations where the variable is raised to a power of \(2\). These are called quadratic equations and are used when launching rockets, maximizing profits, minimizing costs, and making smooth computer animations, to name just a few.
We've seen a few simple quadratic equations, situations where there was a single \(x^2\) and it could be solved by taking a square root. In this section we'll look at equations that have \(x^2\) and \(x\) in them and let factoring help us solve them.
But first we need to establish a very helpful fact.
Solving with the Zero-Product Principle
Suppose we have two quantities, \(A\) and \(B\), and that their product is \(A \cdot B=0\). Does that tell us anything about \(A\) and/or \(B\)?
It does! 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 fact is known as 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. If \(A\cdot B=6\) then there's no guarantee that either \(A=6\) or \(B=6\).
If we encounter an equation where a product equals zero, then we can, with confidence, say that each factor of that product must be zero. Let's walk through a few examples.
Solve \((x+5)(x-7)=0\)
Solution
Applying the zero-product property, we know that either \(x+3=0\) or \(x-7=0\).
\begin{align} && (x+5)(x-& 7) = 0 &&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ x+5 = 0 && \text{ or } && x-7 &= 0 && \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}}\\ x = -5 && \text{ or } && x &= 7 && \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}The two solutions are \(x=-5\) and \(x=7\).
Solve \(2x(3x-4)=0\)
Solution
Applying the zero-product property, we know that either \(2x=0\) or \(3x-4=0\).
\begin{align} && 2x(3x-& 4) = 0 &&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ 2x= 0 && \text{ or } && 3x-4 &= 0 && \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}} \\ x = 0 && \text{ or } && x &= \frac{4}{3} && \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}The two solutions are \(x=0\) and \(x=\frac{4}{3}\).
This process can be extended to more than two factors, as we'll see in the next example.
Solve \((x+2)(x-6)(x-9)=0\)
Solution
Applying the zero-product property, we know that \(x+2=0\) or \(x-6=0\) or \(x-9\).
\begin{align} x+2 = 0 & \quad \text{ or } & x-6 &= 0 & \text{ or } && x-9 = 0 &&& \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}} \\ x = -2 & \quad \text{ or } & x &= 6 & \text{ or } && x = 9 &&& \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}The three solutions are \(x=-2\), \(x=6\), and \(x=9\).
Solving a Quadratic Equation by Factoring
In Examples 1-3, the equations were given to us factored. Most of the time, however, we will need 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-factor 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 &&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ && (x+3)(x-& 2) = 0 &&&& \quad \small \color{#5fa2ce} {\text{Factored equation}}\\ x+3 = 0 && \text{ or } && x-2 &= 0 && \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}}\\ x = -3 && \text{ or } && x &= 2 && \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}The solutions are \(x=-3\) and \(x=2\).
Solve \(x^2-100=0\).
Solution
The right side can be factored using \(10\) and \(-10\).
\begin{align} && x^2 -100 - \, &= 0 &&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ && (x+10)(x-10) &= 0 &&&& \quad \small \color{#5fa2ce} {\text{Factored equation}}\\ x+10 = 0 && \text{ or } && x-10 &= 0 && \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}}\\ x = -10 && \text{ or } && x &= 10 && \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}The solutions are \(x=-10\) and \(x=10\).
Throughout math, whenever we encounter a new problem we always want to turn it into a problem we've solved before. It's no different here. 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, and the proceed as before.
Solve \(4x^2=28x\).
Solution
We'll have to set the equation equal to \(0\) by subtracting the term \(28x\) from both sides of the equation.
\begin{align} && 4x^2 &= 28x &&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ && 4x^2 -28x &= 0 &&&& \quad \small \color{#5fa2ce} {\text{Subtract }28x}\\ && 4x(x -7) &= 0 &&&& \quad \small \color{#5fa2ce} {\text{Factor the GCF of }4x}\\ 4x = 0 && \text{ or } && x-7 &= 0 && \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}}\\ x = 0 && \text{ or } && x &= 7 && \quad \small \color{#5fa2ce} {\text{Solve each equation}} \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 &&&&& \quad \small \color{#5fa2ce} {\text{Original equation}}\\ &&&& 5x(x^2 + 2x -15) &= 0 &&&&& \quad \small \color{#5fa2ce} {\text{Factor the GCF of }5x}\\ &&&& 5x(x+5)(x-3) &= 0 &&&&& \quad \small \color{#5fa2ce} {\text{Factor the trinomial}}\\ \\ 5x &= 0 & \text{ or } && x+5 &= 0 & \text{ or } && x-3 &= 0 & \quad \small \color{#5fa2ce} {\text{Apply Zero-Product Property}}\\ x &= 0 & \text{ or } && x &= -5 & \text{ or } && x &= 3 & \quad \small \color{#5fa2ce} {\text{Solve each equation}} \end{align}
The solutions are \(x=0\), \(x=-5\) 3and \(x=3\).
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.
Conclusion
In the last two sections we've greatly expanded the type and variety of equations we can solve, but we're not done yet. There's one final tool we need to add, and that will come in the next section.