We saw previously that the most basic of equations, linear equations, can be solved by isolating the variable of interest. However, we also discovered that the strategy of isolation can apply to equations that are not linear equations. The second most basic category of equations that students learn in an Algebra course is that of quadratic equations. We will see that the basic strategy for solving quadratic equations — factoring — also applies in many other situations.
The strategy of factoring to solve equations is based on a basic principle of arithmetic, which is called Theorem 0.11 in Section 0.2 of the textbook. This principle has to do with the special power of the number zero. Zero is a very powerful number. It is both the most selfless and most selfish number of all. If you add 0 to any number, you get back that number, \(a+0=a\). However, if you multiply any number by 0, you get back 0, \(a \cdot 0 = 0\). Theorem 0.11 tells us that the only way to multiply two values together to get 0, one of the factors must have been 0. \[A \cdot B = 0 \quad \Leftrightarrow \quad A=0 \hbox{ or } B=0.\]
Factoring is the process of rewriting an expression whose final operation is addition as an equivalent expression whose final operation is multiplication. Factoring is an application of the distributive property, only applied in reverse. We factor to solve equations by applying the following steps:
Example: Solve the equation \(x^2=5x\) for \(x\).
Most students see the \(x\) on both sides of the equation and want to divide it away. This is a mistake and will cause you to miss one of the solutions. Because there are two different terms each having \(x\), and we cannot combine the terms to use isolation as a method, we must use the strategy of factoring.
We begin by moving all terms to one side—my preference is usually to move all terms to the left side of the equation. Once we have 0 alone on the right, we will factor the expression on the left. \[ \begin{align*} x^2+-5x &= 5x+-5x \\ x^2-5x &= 0 \\ x(x-5) &= 0 \end{align*} \] Our new equivalent equation is a product of the factors \(x\) and \(x-5\). We solve the equations that determine when each factor equals 0. The factor \(x\) equals 0 precisely when \(x=0\). (That sounds silly to say, though I am simply trying to emphasize the distinction between the equation and the solution which would be written the same way since \(x\) is already isolated.) The factor \(x-5\) equals 0, solving \(x-5=0\) when \(x=5\) (isolation). In this case, each factor contributed one solution: \[x^2=5x \qquad \Leftrightarrow \qquad x \in \{0,5\}.\]
Example: Solve the equation \(x^2-3x-4=0\) for \(x\).
This equation is already written as an expression equals zero. So we proceed directly to factor the quadratic, using a procedure that many students like to call Reverse FOIL. (If you don't remember how to factor quadratics, follow this link.) \[ \begin{align*} x^2 - 3x - 4 &= 0 \\ (x-4)(x+1) &= 0 \end{align*} \] We now have an equation of the form a product equals zero. We solve for when each factor individually equals zero. \[ x-4 = 0 \qquad \Leftrightarrow \qquad x=4, \qquad \qquad x+1=0 \qquad \Leftrightarrow \qquad x=-1.\] The solution set to \(x^2-3x-4=0\) is the set \(\{-1,4\}\).
The strategy of factoring is also useful for solving equations that involve quotients (division) of expressions. This is actually the purpose behind Theorem 0.17 in Section 0.2 of the textbook. However, if we will just think of division as multiplication by a multiplicative inverse, then we really have nothing new. The factors resulting from the multiplicative inverses never add any extra solutions because a multiplicative inverse can never equal zero. On the other hand, we do need to verify that the solutions from other factors do not lead to division by zero (0 has no multiplicative inverse).
Example: Solve the equation \(\displaystyle \frac{x^2-3x-4}{x+3} = 0\).
We begin by rethinking of the quotient as a product: \[ (x^2-3x-4) \div(x+3) = 0.\] The first factor is \(x^2-3x-4\), and we actually solved the equation \(x^2-3x-4=0\) in the last example to find solutions \(\{-1,4\}\). The second factor is \(\div(x+3)\), which by virtue of being a multiplicative inverse can never equal zero. However, we do need to verify that we did not attempt to divide by zero, meaning we need to verify that \(x+3=0\) does not occur for our two solutions \(\{-1,4\}\). This is quickly verified by recognizing that the solution to \(x+3=0\), which is \(x=-3\), does not appear in our list. So our final solution set is given by: \[\frac{x^2-3x-4}{x+3}=0 \qquad \Leftrightarrow \qquad x \in \{-1,4\}.\]
Example: Solve the equation \(\displaystyle \frac{x^2-4}{x^2+3x+2} = 0\).
We begin by factoring both numerator and denominator of our expression and then rethink of our equation as a product. \[ \begin{align*} \frac{x^2-4}{x^2+3x+2} &= 0 \\ \frac{(x+2)(x-2)}{(x+2)(x+1)} &= 0 \\ (x+2)(x-2)\div(x+2)\div(x+1) &= 0 \end{align*} \] The equation consists of a product of four factors. (Make note of, but do not cancel the multiplicative inverses \((x+2)\) and \(\div(x+2)\)! We will want to keep track of all points that give us problems for the original expression.) The non-multiplicative inverse factors lead to equations \(x+2=0\) and \(x-2=0\), with corresponding solutions \(x=-2\) and \(x=2\). While the two multiplicative inverse factors \(\div(x+1)\) and \(\div(x+2)\) do not contribute any new solutions, they do tell us that \(x=-1\) and \(x=-2\) will lead to division by zero. So the solution \(x=-2\) must be removed from our list of solutions. \[\frac{x^2-4}{x^2+3x+2}=0 \qquad \Leftrightarrow \qquad x=2.\]
At this stage, you are expected to know how to factor quadratics. Section 0.2 introduces rules for factoring the difference of powers, \[x^n-a^n = (x-a)(x^{n-1} + ax^{n-2}+a^2x^{n-3} + \cdots + a^{n-1}),\] with special cases for \(n=2\) and \(n=3\), as well as the sum of cubes, \[x^n+a^n = (x+a)(x^2-ax+a^2).\] It is unlikely that you have used these rules enough to remember them, so you will need to practice. More advanced strategies for factoring polynomials are introduced in Section 4.1 of the textbook, particularly the method of synthetic division.
The strategies for solving equations are closely related to strategies for solving inequalities, which is a major focus in Section 0.3 of the textbook. The next section describes the steps required for using the method of isolation to solve an inequality. This is followed by a section describing how to use factoring to solve inequalities.
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