7.6: Solve by factoring
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When solving linear equations, such as \(2x − 5 = 21\), we can solve by isolating the variable on one side and a number on the other side. However, in this chapter, we have an \(x^2\) term, so if it looks different, then it is different. Hence, we need a new method for solving trinomial equations. One method is using the zero product rule. There are other methods for solving trinomial equations, but that is for a future chapter.
A polynomial equation is any equation that contains a polynomial expression. A trinomial equation is written in the form \[ax^2+bx+c=0,\nonumber\] where \(a,\: b,\: c\) are coefficients, and \(a=\neq 0\).
Zero Product Rule
\[a\cdot b=0\text{ implies }a=0\text{ or }b=0\text{, or both }a=b=0\nonumber\]
The zero product rule states that in order for a product to be zero, then one of its factors must be zero, or even both since \(0\cdot 0 = 0\). Let’s extend this rule into solving for a trinomial equation.
Solve for \(x\): \((2x-3)(5x+1)=0\)
Solution
Using the zero product rule, we know that in order for this product to be equal to zero, then at least one of the factors must be zero:
\[\begin{array}{rl}(2x-3)(5x+1)=0&\text{Set seach factor equal to zero} \\ 2x-3=0\quad\text{or}\quad 5x+1=0&\text{Solve} \\ 2x=3\quad\text{or}\quad 5x=-1 \\ x=\dfrac{3}{2}\quad\text{or}\quad x=-\dfrac{1}{5}&\text{Solution}\end{array}\nonumber\]
Solve by Factoring
Step 1. Write the given equation in the form \(ax^2 + bx + c = 0\).
Step 2. Factor the left side of the equation into a product of factors.
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown.
Step 4. Verify the solution(s).
Solve for \(x\): \(4x^2+x-3=0\)
Solution
Step 1. The equation is already given with zero on the right side. \[4x^2+x-3=0\nonumber\]
Step 2. Factor the left side of the equation into a product of factors: \[\begin{aligned}4x^2+x-3&=0 \\ 4x^2-3x+4x-3&=0 \\ x(4x-3)+1(4x-3)&=0 \\ (4x-3)(x+1)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}(4x-3)(x+1)=0&\text{Set each factor equal to zero} \\ 4x-3=0\quad\text{or}\quad x+1=0&\text{Solve} \\ 4x=3\quad\text{or}\quad x=-1 \\ x=\dfrac{3}{4}\quad\text{or}\quad x=-1&\text{Solution}\end{array}\nonumber\]
Step 4. Verify the solution(s): \(x =\dfrac{3}{4}\) and \(x = −1\) \[\begin{array}{rl} 4\color{blue}{\left(\dfrac{3}{4}\right)^2}\color{black}{}+\color{blue}{\left(\dfrac{3}{4}\right)}\color{black}{}-3\stackrel{?}{=}0 & 4\color{blue}{(-1)}\color{black}{}^2+\color{blue}{(-1)}\color{black}{}-3\stackrel{?}{=}0 \\ 4\cdot\dfrac{9}{16}+\dfrac{3}{4}-3\stackrel{?}{=}0&4(1)-1-3\stackrel{?}{=}0 \\ 0=0\quad\checkmark & 0=0\quad\checkmark\end{array}\nonumber\]
Thus, the solutions are \(x=\dfrac{3}{4}\) and \(x=-1\).
Rewrite the Equation with Zero on One Side
Solve for \(x\): \(x^2=8x-15\)
Solution
Step 1. Write the given equation in the form with zero on the right side: \[\begin{aligned}x^2&=8x-15 \\ x^2-8x+15&=0\end{aligned}\]
Step 2. Factor the left side of the equation into a product of factors: \[\begin{aligned} x^2-8x+15&=0 \\ (x-5)(x-3)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}(x-5)(x-3)=0&\text{Set each factor equal to zero} \\ x-5=0\quad\text{or}\quad x-3=0&\text{Solve} \\ x=5\quad\text{or}\quad x=3&\text{Solution}\end{array}\nonumber\]
Step 4. Verify the solution(s): \(x = 5\) and \(x = 3\) \[\begin{array}{rl}\color{blue}{(5)}\color{black}{}^2\stackrel{?}{=}8\color{blue}{(5)}\color{black}{}-15&\color{blue}{(3)}\color{black}{}^2\stackrel{?}{=}8\color{blue}{(3)}\color{black}{}-15 \\ 25\stackrel{?}{=}40-15&9\stackrel{?}{=}24-15 \\ 25=25\quad\checkmark &9=9\quad\checkmark\end{array}\nonumber\]
Thus, the solutions are \(x=5\) and \(x=3\).
Simplify the Equation
Sometimes the equation isn’t so straightforward. We may have to do some preliminary work so that the equation takes the form of a trinomial equation and then we can use the zero product rule.
Solve for \(x\): \((x-7)(x+3)=-9\)
Solution
Step 1. Write the given equation in the form with zero on the right side. Notice, we will have to FOIL the left side first, then obtain zero on the right. \[\begin{aligned}(x-7)(x+3)&=-9 \\ x^2-4x-21&=-9 \\ x^2-4x-12&=0\end{aligned}\]
Step 2. Factor the left side of the equation into a product of factors: \[\begin{aligned}x^2-4x-12&=0 \\ (x-6)(x+2)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}(x-6)(x+2)=0&\text{Set each factor equal to zero} \\ x-6=0\quad\text{or}\quad x+2=0&\text{Solve} \\ x=6\quad\text{or}\quad x=-2&\text{Solution}\end{array}\nonumber\]
Step 4. Verify the solution(s): \(x = 6\) and \(x = −2\) \[\begin{array}{rl}(\color{blue}{(6)}\color{black}{}-7)(\color{blue}{(6)}\color{black}{}+3)\stackrel{?}{=}-9&(\color{blue}{(-2)}\color{black}{}-7)(\color{blue}{(-2)}\color{black}{}+3)\stackrel{?}{=}-9 \\ -9=-9\quad\checkmark &-9=-9\quad\checkmark\end{array}\nonumber\]
Thus, the solutions are \(x=6\) and \(x=-2\).
Solve for \(x\): \(3x^2+4x-5=7x^2+4x-14\)
Solution
Step 1. Write the given equation in the form with zero on the right side. Notice, we will have to combine like terms to obtain zero on the right. \[\begin{aligned}3x^2+4x-5&=7x^2+4x-14 \\ -4x^2+9&=0 \\ \color{blue}{(-1)}\color{black}{}(-4x^2+9)&=0\color{blue}{(-1)}\color{black}{} \\ 4x^2-9&=0\end{aligned}\]
Step 2. Factor the left side of the equation into a product of factors: \[\begin{aligned}4x^2-9&=0 \\ (2x+3)(2x-3)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}(2x+3)(2x-3)=0&\text{Set each factor equal to zero} \\ 2x+3=0\quad\text{or}\quad 2x-3=0 &\text{Solve} \\ x=-\dfrac{3}{2}\quad\text{or}\quad x=\dfrac{3}{2}&\text{Solution}\end{array}\nonumber\]
Step 4. We leave verifying the solution(s): \(x = −\dfrac{3}{2}\) and \(x =\dfrac{3}{2}\), to the student.
Solve for \(x\): \(4x^2=12x-9\)
Solution
Step 1. Write the given equation in the form with zero on the right side: \[\begin{aligned}4x^2&=12x-9 \\ 4x^2-12x+9&=0\end{aligned}\]
Step 2. Factor the left side of the equation into a product of factors: \[\begin{aligned}4x^2-12x+9&=0 \\ (2x-3)^2&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}(2x-3)^2=0&\text{Rewrite as two factors} \\ (2x-3)(2x-3)=0&\text{Set each factor equal to zero} \\ 2x-3=0\quad\text{or}\quad 2x-3=0&\text{Solve} \\ x=\dfrac{3}{2}\quad\text{or}\quad x=\dfrac{3}{2}&\text{Solution}\end{array}\nonumber\] Notice we obtain the same solution for both factors. Even though we usually obtain two different solutions, in some cases, we obtain one solution. We call this solution with multiplicity two.
Step 4. We leave verifying the solution(s): \(x =\dfrac{3}{2}\), to the student.
Thus, the solution is \(x=\dfrac{3}{2}\) with multiplicity two.
In solving trinomials of the form \(ax^2 + bx + c = 0\), we should always obtain two solutions. There is one case in which we will obtain one solution with multiplicity two. This case is when the trinomial equation is a perfect square trinomial.
Solve for \(x\): \(4x^2=8x\)
Solution
Step 1. Write the given equation in the form with zero on the right side. \[\begin{aligned}4x^2&=8x \\ 4x^2-8x&=0\end{aligned}\]
Step 2. Factor the left side of the equation into a product of factors. Notice here, we will only factor a GCF. \[\begin{aligned}4x^2-8x&=0 \\ 4x(x-2)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}4x(x-2)=0&\text{Set each factor equal to zero} \\ 4x=0\quad\text{or}\quad x-2=0&\text{Solve} \\ x=0\quad\text{or}\quad x=2&\text{Solution}\end{array}\nonumber\]
Step 4. We leave verifying the solution(s): \(x = 0\) and \(x = 2\), to the student.
Solve for \(x\): \(2x^3-14x^2+24x=0\)
Solution
Step 1. We were given the equation in the form with zero on the right side: \[2x^3-14x^2+24x=0\nonumber\]
Step 2. Factor the left side of the equation into a product of factors. Notice here, we will factor a GCF in addition to factoring the trinomial. \[\begin{aligned}2x^3-14x^2+24x&=0 \\ 2x(x^2-7x+12)&=0 \\ 2x(x-3)(x-4)&=0\end{aligned}\]
Step 3. Use the zero product rule to set each factor equal to zero and then solve for the unknown: \[\begin{array}{rl}2x(x-3)(x-4)=0&\text{Set each factor equal to zero} \\ 2x=0\quad\text{or}\quad x-3=0\quad\text{or}\quad x-4=0&\text{Solve} \\ x=0\quad\text{or}\quad x=3\quad\text{or}\quad x=4&\text{Solution}\end{array}\nonumber\]
Step 4. We leave verifying the solution(s): \(x = 0,\: x = 3\), and \(x = 2\), to the student.
Notice, we obtained three solutions to the equation. Although we were given a trinomial, notice the degree of the trinomial was \(3\), i.e., when we factored, we obtained three factors. Hence, we will have three solutions. In general, the number of solutions will be at most the number of factors, e.g., we obtain two factors and one solution with multiplicity two.
While factoring works great to solve problems with an \(x^2\) term, Tartaglia, in \(16^{\text{th}}\) century Italy, developed a method to solve problems with \(x^3\). He kept his method a secret until another mathematician, Cardan, talked him out of his secret and published the results. To this day, the formula is known as Cardan’s Formula
Solve by Factoring Homework
Solve each equation by factoring.
\((k-7)(k+2)=0\)
\((x-1)(x+4)=0\)
\(6x^2-150=0\)
\(2n^2+10n-28=0\)
\(7x^2+26x+15=0\)
\(5n^2-9n-2=0\)
\(x^2-4x-8=-8\)
\(x^2 − 5x − 1 = −5\)
\(49p^2 + 371p − 163 = 5\)
\(7x^2 + 17x − 20 = −8\)
\(7r^2 + 84 = −49r\)
\(x^2 − 6x = 16\)
\(3v^2 + 7v = 40\)
\(35x^2 + 120x = −45\)
\(4k^2 + 18k − 23 = 6k − 7\)
\(9x^2 − 46 + 7x = 7x + 8x^2 + 3\)
\(2m^2 + 19m + 40 = −2m\)
\(40p^2 + 183p − 168 = p + 5p^2\)
\((a + 4)(a − 3) = 0\)
\((2x + 5)(x − 7) = 0\)
\(p^2 + 4p − 32 = 0\)
\(m^2 − m − 30 = 0\)
\(40r^2 − 285r − 280 = 0\)
\(2b^2 − 3b − 2 = 0\)
\(v^2 − 8v − 3 = −3\)
\(a^2 − 6a + 6 = −2\)
\(7k^2 + 57k + 13 = 5\)
\(4n^2 − 13n + 8 = 5\)
\(7m^2 − 224 = 28m\)
\(7n^2 − 28n = 0\)
\(6b^2 = 5 + 7b\)
\(9n^2 + 39n = −36\)
\(a^2 + 7a − 9 = −3 + 6a\)
\(x^2 + 10x + 30 = 6\)
\(5n^2 + 41n + 40 = −2\)
\(24x^2 + 11x − 80 = 3x\)
\((x − 3)^2 − 6(x − 3) + 8 = 0\)
\(4(2 − 5x)^2 + 8(2 − 5x) = −3\)