Solving Quadratic Equations by Factoring
Learning how to solve equations is one of our main goals in algebra. Up to this point, we have solved linear equations, which are of degree 1. In this section, we will learn a technique that can be used to solve certain equations of degree 2. A quadratic equation is any equation that can be written in the standard form
\[ax^{2}+bx+c=0\],
where \(a, b\), and \(c\) are real numbers and \(a≠0\). The following are some examples of quadratic equations, all of which will be solved in this section:
\(x^{2}+x-6=0\)
\(4x^{2}-9=0\)
\(2x^{2}+10x+20=-3x+5\)
A solution of a quadratic equation in standard form is called a root. Quadratic equations can have two real solutions, one real solution, or no real solution. The quadratic equation \(x^{2}+x−6=0\) has two solutions, namely, \(x=−3\) and \(x=2\).
Example \(\PageIndex{1}\)
Verify that \(x=−3\) and \(x=2\) are solutions to \(x^{2}+x−6=0\).
Solution:
To verify solutions, substitute the values for \(x\) and then simplify to see if a true statement results.
\(\begin{array} {r|r} {\color{Cerulean}{Check:}\color{black}{\:x=-3}}&{\color{Cerulean}{Check:}\color{black}{\:x=2}} \\ {x^{2}+x-6=0}&{x^{2}+x-6=0}\\{(\color{OliveGreen}{-3}\color{black}{)^{2}+(}\color{OliveGreen}{-3}\color{black}{)-6=0}}&{(\color{OliveGreen}{2}\color{black}{)^{2}+(}\color{OliveGreen}{2}\color{black}{)-6=0}}\\{9-3-6=0}&{4+2-6=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)
Answer:
Both values produce true statements. Therefore, they are both solutions to the equation.
Our goal is to develop algebraic techniques for finding solutions to quadratic equations. The first technique requires the zero-product property:
If \(a\cdot b=0\), then \(a=0\) or \(b=0\)
In other words, if any product is equal to zero, then one or both of the variable factors must be equal to zero.
Example \(\PageIndex{2}\)
Solve:
\((x−8)(x+7)=0\).
Solution:
This equation consists of a product of two quantities equal to zero; therefore, the zero-product property applies. One or both of the quantities must be zero.
\(\begin{array} {ccc} {(x-8)=0}&{\text{or}}&{(x+7)=0}\\{x-8\color{Cerulean}{+8}\color{black}{=0}\color{Cerulean}{+8}}&{}&{x+7\color{Cerulean}{-7}\color{black}{=0}\color{Cerulean}{-7}}\\{x=8}&{}&{x=-7} \end{array}\)
To verify that these are solutions, substitute them for the variable \(x\).
\(\begin{array}{r|r} {\color{Cerulean}{Check:}\color{black}{\:x=8}}&{\color{Cerulean}{Check:}\color{black}{\:x=-7}}\\{(x-8)(x+7)=0}&{(x-8)(x+7)=0}\\{(8-8)(8+7)=0}&{(-7+8)(-7+7)=0}\\{(0)(15)=0}&{(1)(0)=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)
Notice that each solution produces a factor that is equal to zero.
Answer:
The solutions are \(8\) and \(−7\).
The quadratic equation may not be given in its factored form.
Example \(\PageIndex{3}\)
Solve:
\(x^{2}+3x−10=0\).
Solution:
The goal is to produce a product that is equal to zero. We can do that by factoring the trinomial on the left side of the equation.
\(\begin{array}{cc}{x^{2}+3x-10=0}&{}\\{(x\quad\color{Cerulean}{?}\color{black}{)(x\quad}\color{Cerulean}{?}\color{black}{)=0}}&{\color{Cerulean}{-10=5(-2)}}\\{}&{\color{Cerulean}{and\: 3=5+(-2)}}\\{(x+5)(x-2)=0}&{} \end{array}\)
Next, apply the zero-product property and set each factor equal to zero.
\(x+5=0 \quad\text{or}\quad x-2=0\)
This leaves us with two linear equations, each of which can be solved for \(x\).
\(\begin{array}{cc} {x+5\color{Cerulean}{-5}\color{black}{=0}\color{Cerulean}{-5}}&{x-2\color{Cerulean}{+2}\color{black}{=0}\color{Cerulean}{+2}}\\{x=-5}&{x=2} \end{array}\)
Check the solutions by substituting into the original equation to verify that we obtain true statements.
\(\begin{array}{r|r}{\color{Cerulean}{Check:}\color{black}{\:x=-5}}&{\color{Cerulean}{Check:}\color{black}{\:x=2}}\\{x^{2}+3x-10=0}&{x^{2}+3x-10=0}\\{(\color{OliveGreen}{-5}\color{black}{)^{2}+3(}\color{OliveGreen}{-5}\color{black}{)-10=0}}&{(\color{OliveGreen}{2}\color{black}{)^{2}+3(}\color{OliveGreen}{2}\color{black}{)-10=0}}\\{25-15-10=0}&{4+6-10=0}\\{0=0\quad\color{Cerulean}{\checkmark}}&{0=0\quad\color{Cerulean}{\checkmark}} \end{array}\)
Answer:
The solutions are \(-5\) and \(2\).
Using the zero-product property after factoring a quadratic equation in standard form is the key to this technique. However, the quadratic equation may not be given in standard form, and so there may be some preliminary steps before factoring. The steps required to solve by factoring are outlined in the following example.
Example \(\PageIndex{4}\)
Solve:
\(2x^{2}+10x+20=−3x+5\).
Solution:
Step 1: Express the quadratic equation in standard form. For the zero-product property to apply, the quadratic expression must be equal to zero. Use the addition and subtraction properties of equality to combine opposite-side like terms and obtain zero on one side of the equation. In this example, add \(3x\) to and subtract \(5\) from both sides.
\(\begin{aligned} 2x^{2}+10x+20\color{Cerulean}{+3x}&\color{black}{=-3x+5}\color{Cerulean}{+3x} \\ 2x^{2}+13x+20&=5 \\ 2x^{2}+13x+20\color{Cerulean}{-5}&\color{black}{=5}\color{Cerulean}{-5} \\ 2x^{2}+13x+15&=0 \end{aligned}\)
Step 2: Factor the quadratic expression.
\((2x+3)(x+5)=0\)
Step 3: Apply the zero-product property and set each variable factor equal to zero.
\(2x+3=0\quad\text{or}\quad x+5=0\)
Step 4: Solve the resulting linear equations.
\(\begin{array} {ccc} {2x+3=0}&{\text{or}}&{x+5=0}\\{2x=-3}&{}&{x=-5}\\{\frac{2x}{\color{Cerulean}{2}}\color{black}{=\frac{-3}{\color{Cerulean}{2}}}}&{}&{}\\{x=-\frac{3}{2}}&{}&{} \end{array}\)
Answer:
The solutions are \(-5\) and \(-\frac{3}{2}\). The check is optional.
Example \(\PageIndex{5}\)
Solve:
\(9x^{2}+1=6x\).
Solution:
Write this in standard form by subtracting \(6x\ from both sides.
\(\begin{aligned} 9x^{2}+1\color{Cerulean}{-6x}&\color{black}{=6x}\color{Cerulean}{-6x} \\ 9x^{2}-6x+1&=0 \end{aligned}\)
Once the equation is in standard form, equal to zero, factor.
\((3x-1)(3x-1)=0\)
This is a perfect square trinomial. Hence setting each factor equal to zero results in a repeated solution.
\(\begin{array}{ccc}{3x-1=0}&{\text{or}}&{3x-1=0}\\{3x=1}&{}&{3x=1}\\{x=\frac{1}{3}}&{}&{x=\frac{1}{3}}\\ \end{array}\)
A repeated solution is called a double root and does not have to be written twice.
Answer:
The solution is \(\frac{1}{3}\).
Exercise \(\PageIndex{1}\)
Solve:
\(x^{2}−3x=28\).
- Answer
-
\(x=−4\) or \(x=7\)
Not all quadratic equations in standard form are trinomials. We often encounter binomials.
Example \(\PageIndex{6}\)
Solve:
\(x^{2}−9=0\).
Solution:
This quadratic equation is given in standard form, where the binomial on the left side is a difference of squares. Factor as follows:
\(\begin{aligned} x^{2}-9&=0 \\ (x+3)(x-3)&=0 \end{aligned}\)
Next, set each factor equal to zero and solve.
\(\begin{array}{ccc}{x+3=0}&{\text{or}}&{x-3=0}\\{x=-3}&{}&{x=3}\end{array}\)
Answer:
The solutions are \(3\) and \(−3\), which can also be written as \(±3\).
Example \(\PageIndex{7}\)
Solve:
\(5x^{2}=15x\)
Solution:
By inspection, we see that \(x=0\) is a solution to this quadratic equation. Since dividing by zero is undefined, we want to avoid dividing both sides of this equation by \(x\). In general, we wish to avoid dividing both sides of any equation by a variable or an expression containing a variable. We will discuss this in more detail later. The first step is to rewrite this equation in standard form with zero on one side.
\(\begin{aligned} 5x^{2}&=15x \\ 5x^{2}\color{Cerulean}{-15x}&\color{black}{=15x}\color{Cerulean}{-15x\quad Subtract\:15x\:from\:both\:sides.} \\ 5x^{2}-15x&=0 \end{aligned}\)
Next, factor the expression. Notice that the binomial on the left has a GCF of \(5x\).
\(\begin{aligned} 5x^{2}-15x&=0\\5x(x-3)&=0 \end{aligned}\)
Set each factor to equal to zero.
\(\begin{array}{ccc}{5x=0}&{\text{or}}&{x-3=0}\\{\frac{5x}{\color{Cerulean}{5}}\color{black}{=\frac{0}{\color{Cerulean}{5}}}}&{}&{x=3}\\{x=0}&{}&{} \end{array}\)
Answer:
The solutions are \(0\) and \(3\).
Example \(\PageIndex{8}\)
Solve:
\((2x+1)(x+5)=11\).
Solution:
This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(11\). However, this would lead to incorrect results. We must rewrite the equation in standard form, equal to zero, so that we can apply the zero-product property.
\(\begin{aligned}(2x+1)(x+5)&=11 \\ 2x^{2}+10x+x+5&=11 \\ 2x^{2}+11x+5\color{Cerulean}{-11}&\color{black}{=11}\color{Cerulean}{-11}\\2x^{2}+11x-6&=0 \end{aligned}\)
Once it is in standard form, we can factor and then set each factor equal to zero.
\((2x-1)(x+6)=0\)
\(\begin{array}{ccc}{2x-1=0}&{\text{or}}&{x+6=0}\\{2x=1}&{}&{x=-6}\\{x=\frac{1}{2}}&{}&{} \end{array}\)
Answer:
The solutions are \(\frac{1}{2}\) and \(-6\).
Example \(\PageIndex{9}\)
Solve:
\(15x^{2}−25x+10=0\).
Solution:
We begin by factoring out the GCF of \(5\). Then factor the resulting trinomial.
\(\begin{aligned} 15x^{2}-25x+10&=0 \\ 5(3x^{2}-5x+2)&=0\\5(3x-2)(x-1)&=0 \end{aligned}\)
Next, we set each variable factor equal to zero and solve for \(x\).
\(\begin{array}{ccc} {3x-2=0}&{\text{or}}&{x-1=0}\\{3x=2}&{}&{x=1}\\{x=\frac{2}{3}}&{}&{} \end{array}\)
Notice that the factor \(5\) is not a variable factor and thus did not contribute to the solution set.
Answer:
The solutions are \(\frac{2}{3}\) and \(1\).
Example \(\PageIndex{10}\)
Factor:
\(52x^{2}+76x−13=0\).
Solution:
Clear the fractions by multiplying both sides of the equation by the LCD, which is equal to \(6\).
\(\begin{aligned} \frac{5}{2}x^{2}+\frac{7}{6}x-\frac{1}{3}&=0 \\ \color{Cerulean}{6}\color{black}{\cdot} \left( \frac{5}{2}x^{2}+\frac{7}{6}x-\frac{1}{3} \right) &=\color{Cerulean}{6}\color{black}{\cdot (0)} \\ 15x^{2}+7x-2&=0 \end{aligned}\)
At this point, we have an equivalent equation with integer coefficients and can factor as usual. Begin with the factors of \(15\) and \(2\).
\(\begin{array}{cc}{15=1\cdot 15}&{2=1\cdot 2} \\ {=3\cdot 5}&{} \end{array}\)
The coefficient of the middle term is \(7=3(−1)+5(2)\). Factor as follows:
\(\begin{aligned} 15x^{2}+7x-2&=0 \\ (3x+2)(5x-1)&=0 \end{aligned}\)
Set each factor equal to zero and solve.
\(\begin{array}{ccc}{3x+2=0}&{\text{or}}&{5x-1=0}\\{3x=-2}&{}&{5x=1}\\{x=-\frac{2}{3}}&{}&{x=\frac{1}{5}} \end{array}\)
Answer:
The solutions are \(-\frac{2}{3}\) and \(\frac{1}{5}\).
Exercise \(\PageIndex{2}\)
Solve:
\(4x^{2}-9=0\).
- Answer
-
\(-\frac{3}{2}\) and \(\frac{3}{2}\)
Finding Equations with Given Solutions
The zero-product property states,
If \(a\cdot b=0\), then \(a=0\) or \(b=0\)
And, in fact, the converse is true as well:
If \(a=0\) or \(b=0\), then \(ab=0\)
When this is the case, we can write the following:
\(a\cdot b =0\), if and only if \(a=0\) or \(b=0\)
We use this property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.
Example \(\PageIndex{11}\)
Find a quadratic equation with solutions \(−7\) and \(2\).
Solution:
Given the solutions, we can determine two linear factors.
\(\begin{array} {ccc}{x=-7}&{\text{or}}&{x=2}\\{x+7=0}&{}&{x-2=0} \end{array}\)
The product of these linear factors is equal to zero when \(x=-7\) or \(x=2\):
\((x+7)(x-2)=0\)
Multiply the binomials and present the equation in standard form.
\(\begin{aligned} x^{2}-2x+7x-14&=0 \\ x^{2}+5x-14&=0 \end{aligned}\)
Answer:
\(x^{2}+5x-14=0\).
We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found aboveis not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.
Example \(\PageIndex{12}\)
Find a quadratic equation with integer coefficients, given solutions \(\frac{1}{2}\) and \(−\frac{3}{4}\).
Solution:
To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.
\(\begin{array} {ccc}{x=\frac{1}{2}}&{\text{or}}&{x=-\frac{3}{4}}\\{\color{Cerulean}{2}\color{black}{\cdot x =}\color{Cerulean}{2}\color{black}{\cdot \frac{1}{2}}}&{}&{\color{Cerulean}{4}\color{black}{\cdot x = -\frac{3}{4}\cdot}\color{Cerulean}{4}}\\{2x=1}&{}&{4x=-3}\\{2x-1=0}&{}&{4x+3=0} \end{array}\)
Apply the zero-product property and multiply.
\(\begin{aligned} (2x-1)(4x+3)&=0 \\ 8x^{2}+6x-4x-3&=0 \\ 8x^{2}+2x-3&=0 \end{aligned}\)
Answer:
\(8x^{2}+2x-3=0\)
Exercise \(\PageIndex{3}\)
Find a quadratic equation with integer coefficients, given solutions \(−1\) and \(\frac{2}{3}\).
- Answer
-
\(3x^{2}+x-2=0\)
Solving Polynomial Equations by Factoring
The zero-product property is true for any number of factors that make up an equation. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.
Example \(\PageIndex{13}\)
Solve:
\(3x(x−5)(3x−2)=0\).
Solution:
Set each variable factor equal to zero and solve.
\(\begin{array}{ccccc} {3x=0}&{\text{or}}&{x-5=0}&{\text{or}}&{3x-2=0}\\{\frac{3x}{\color{Cerulean}{3}}\color{black}{=\frac{0}{\color{Cerulean}{3}}}}&{}&{x=5}&{}&{\frac{3x}{\color{Cerulean}{3}}\color{black}{=\frac{2}{\color{Cerulean}{3}}}}\\{x=0}&{}&{}&{}&{x=\frac{2}{3}}\end{array}\)
Answer:
The solutions are \(0, 5\), and \(\frac{2}{3}\).
Of course, we cannot expect the equation to be given in factored form.
Example \(\PageIndex{14}\)
Solve:
\(x^{3}+2x^{2}−9x−18=0\).
Solution
Begin by factoring the left side completely.
\(\begin{array}{rl} {x^{3}+2x^{2}-9x-18=0}&{\color{Cerulean}{Factor\:by\:grouping.}}\\{x^{2}(x+2)-9(x+2)=0}&{}\\{(x+2)(x^{2}-9)=0}&{\color{Cerulean}{Factor\:as\:a\:difference\:of\:squares.}}\\{(x+2)(x+3)(x-3)=0}&{} \end{array}\)
Set each factor equal to zero and solve.
\(\begin{array} {ccccc}{x+2=0}&{\text{or}}&{x+3=0}&{\text{or}}&{x-3=0}\\{x=-2}&{}&{x=-3}&{}&{x=3} \end{array}\)
Answer:
The solutions are \(-2, -3\), and \(3\).
Notice that the degree of the polynomial is \(3\) and we obtained three solutions. In general, for any polynomial equation with one variable of degree \(n\), the fundamental theorem of algebra guarantees \(n\) real solutions or fewer. We have seen that many polynomials do not factor. This does not imply that equations involving these unfactorable polynomials do not have real solutions. In fact, many polynomial equations that do not factor do have real solutions. We will learn how to solve these types of equations as we continue in our study of algebra.
Exercise \(\PageIndex{4}\)
Solve:
\(−10x^{3}−18x^{2}+4x=0\).
- Answer
-
\(−2, 0, \frac{1}{5}\)
