Solving Equations Using Roots and Factoring

Learning Objectives

Polynomial Equations

Solve equations with exponents using the even root property and odd root property
Use factoring methods to factor polynomial equations
Use the principle of zero products to solve polynomial equations

Solving Equations Using Roots

Another type of equation we can solve is one with exponents. As you might expect, we can clear exponents by using roots. This is done with very few unexpected results when the exponent is odd. We can solve these problems in a very straight forward way using the odd root property.

Odd Root Property: if $a^n=b$ and n is odd, then $a={\sqrt[n]{b}}$

Example

Solve the equation: $x^5=32$

Show Solution

$\begin{align}x^5=32&&& \text{Use odd root property}\\x=\sqrt[5]{32}&&& \text{Simplify root}\\x=2&&&\text{Our solution}\end{align}$

However, when the exponent is even we will have two results from taking an even root of both sides. One will be positive and one will be negative. This is because both $3^2 = 9$ and $( − 3)^2 = 9$ . So when solving $x^2 = 9$ we will have two solutions, one positive and one negative: $x = 3$ and $− 3$ .

Even Root Property: if $a^n=b$ and n is even, then $a=\pm \sqrt[n]{b}$

Example

Solve the equation: $x^4=16$

Show Solution

$\begin{align}x^4=16&&& \text{Use even root property}(\pm)\\x=\pm\sqrt[4]{16}&&& \text{Simplify root}\\x=\pm 2&&&\text{Our solution}\end{align}$

World View Note: In 1545, French Mathematician Gerolamo Cardano published his book The Great Art, or the Rules of Algebra which included the solution of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions!

Example

Solve the equation: $(2x+4)^2=36$

Show Solution

$\begin{align}(2x+4)^2=36&&& \text{Use even root property}(\pm)\\2x+4=\pm\sqrt{36}&&& \text{Simplify root}\\2x+4=\pm 6&&&\text{To avoid sign errors we need two equations}\\2x+4=6\text{ or }2x+4=-6&&&\text{One equation for -, one equation for +}\\\underline{-4\hspace{2 mm}-4}\hspace{15 mm}\underline{-4\hspace{2 mm}-4}&&&\text{Subtract 4 from both sides}\\2x=2\text{ or }2x=-10&&&\text{Divide both sides by 2}\\x=1\text{ or }x=-5&&&\text{Our solutions}\end{align}$

In the previous example we needed two equations to simplify because when we took the root, our solutions were two rational numbers, 6 and − 6. If the roots did not simplify to rational numbers we can keep the ± in the equation.

Example

Solve the equation: $(6x-9)^2=45$

Show Solution

$\begin{align}(6x-9)^2=45&&& \text{Use even root property}(\pm) \\6x-9=\pm \sqrt{45}&&& \text{Simplify root}\\ 2x+4=\pm 3\sqrt{5}&&&\text{Use one equation because root did not simplify to rational}\\ \underline{+9\hspace{4 mm}+9}\hspace{5 mm}&&&\text{Add 9 to both sides}\\6x=9\pm 3\sqrt{5}&&&\text{Divide both sides by 6}\\ x=\frac{9\pm 3\sqrt{5}}{6}&&&\text{Simplify, divide each term by 3}\\x=\frac{3\pm \sqrt{5}}{2}&&&\text{Our solutions}\end{align}$

When solving with exponents, it is important to first isolate the part with the exponent before taking any roots.

Example

Solve the equation: $(x+4)^3-6=119$

Show Solution

$\begin{align}(x+4)^3-6=119&&& \text{Isolate the part with the exponent}\\ \underline{+6\hspace{4mm}+6}&&&\text{Add 6 to both sides}\\(x+4)^3=125&&&\text{Use odd root property}(\pm)\\x+4=\sqrt[3]{125}&&& \text{Simplify root}\\x+4=5&&&\\\underline{-4\hspace{2 mm}-4}\\x=1&&&\text{Our solution}\end{align}$

Example

Solve the equation: $(6x+1)^2+6=10$

Show Solution

$\begin{align}(6x+1)^2+6=10&&& \text{Isolate the part with the exponent}\\ \underline{-6\hspace{2mm}-6}&&&\text{Subtract 6 from both sides}\\(6x+1)^2=4&&& \text{Use even root property}(\pm)\\6x+1=\pm\sqrt{4}&&& \text{Simplify root}\\6x+1=\pm 2&&&\text{To avoid sign errors we need two equations}\\6x+1=2\text{ or }6x+1=-2&&&\text{Solve each equation}\\\underline{-1\hspace{2 mm}-1}\hspace{15 mm}\underline{-1\hspace{2 mm}-1}&&&\text{Subtract 1 from both sides}\\6x=1\text{ or }6x=-3&&&\text{Divide both sides by 6}\\\overline{6}\hspace{8mm}\overline{6}\hspace{1mm}\text{ or }\overline{6}\hspace{10mm}\overline{6}\\x=\frac{1}{6}\text{ or }x=-\frac{1}{2}&&&\text{Our solutions}\end{align}$

Solving Equations by Factoring

When solving linear equations or equations with only one power of x we can solve the equation by isolating the variable. However, when we have more than one power of x , such as x and x², we cannot solve in the same way. One method that we can use to solve for the variable is known as the zero product rule.

Zero Product Rule: If ab=0, then either a=0 or b=0

The zero product rule tells us that if two factors are multiplied together and the answer is zero, then one of the factors must be zero. We can use this to help us solve factored polynomials as in the following example.

Example

Solve $(2x-3)(5x+1)=0$

Show Solution

$\begin{align}(2x-3)(5x+1)=0&&&\text{One factor must be zero}\\2x-3=0\text{ or }5x+1=0&&&\text{Set each factor equal to zero and solve}\end{align}$

$\begin{align}2x-3=0\\x=\frac{3}{2}\end{align}$ or $\begin{align}5x+1=0\\x=-\frac{1}{5}\end{align}$

For the zero product rule to work we must have factors to set equal to zero. This means if the problem is not already factored we will factor it first.

Example

Solve $4x^2+x-3=0$

Show Solution

$\begin{align}4x^2+x-3=0&&&\text{Factor using the ac method, multiply to −12, add to 1}\\4x^2-3x+4x-3=0&&&\text{The numbers are -3 and 4, split the middle term}\\x(4x-3)+1(4x-3)=0&&&\text{Factor by grouping}\\(4x-3)(x+1)=0&&&\text{One factor must be 0, set each equal to 0 and solve}\end{align}$

$\begin{align}4x-3=0\\x=\frac{3}{4}\end{align}$ or $\begin{align}x+1=0\\x=-1\end{align}$

Another important part of the zero product rule is that before we factor, the equation must equal zero. If it does not, we must move terms around so it does equal zero. Generally we like the x² term to be positive.

Example

Solve $x^2=8x-15$

Show Solution

$\begin{align}x^2&=8x-15&&\text{Set equal to 0, move 8x and -15 to the left hand side}\\x^2-8x+15&=0&&\text{Factor the left hand side}\\(x-5)(x-3)&=0&&\text{One factor must be 0, set each equal to 0 and solve}\end{align}$

$\begin{align}x-5=0\\x=5\end{align}$ or $\begin{align}x-3=0\\x=3\end{align}$

Put Answer Here

Example

Solve $(x-7)(x+3)=-9$

Show Solution

$\begin{align}(x-7)(x+3)&=-9&&\text{Not equal to 0, multiply parentheses on left hand side}\\x^2-4x-21&=-9&&\text{Set equal to 0, move -9 to the left hand side}\\x^2-4x-12&=0&&\text{Factor the left hand side}\\(x-6)(x+2)&=0&&\text{Set each factor equal to 0 and solve}\end{align}$

$\begin{align}x-6=0\\x=6\end{align}$ or $\begin{align}x+2=0\\x=-2\end{align}$

Example

Solve $3x^2+4x-5=7x^2+4x-14$

Show Solution

$\begin{align}3x^2+4x-5&=7x^2+4x-14&&\text{Set equal to 0 by moving terms to the right hand side}\\0&=4x^2-9&&\text{Factor using difference of squares}\\0&=(2x+3)(2x-3)&&\text{Set each factor equal to 0 and solve}\end{align}$

$\begin{align}2x+3&=0\\x&=-\frac{3}{2}\end{align}$ or $\begin{align}2x-3&=0\\x&=\frac{3}{2}\end{align}$

Most problems with x² will have two unique solutions. However, it is possible to have only one solution as the next example illustrates.

Example

Solve $4x^2=12x-9$

Show Solution

$\begin{align}4x^2&=12x-9&&\text{Set equal to 0 by moving terms to the left hand side}\\4x^2-12x+9&=0&&\text{Factor the left hand side}\\(2x-3)(2x-3)=0\text{ or }(2x-3)^2&=0&&\text{There is one repeated factor. Set it equal to 0 and solve}\end{align}$

$\begin{align}2x-3&=0\\x&=\frac{3}{2}\end{align}$

As always it will be important to factor out the GCF first if we have one. This GCF is also a factor and must also be set equal to zero using the zero product rule. This may give us more than just two solution. The next few examples illustrate this.

Example

Solve $4x^2=8x$

Show Solution

$\begin{align}4x^2&=8x&&\text{Set equal to 0 by moving terms to the left hand side}\\4x^2-8x&=0&&\text{Factor the GCF out of left hand side}\\4x(x-2)&=0&&\text{Set each factor equal to 0 and solve}\end{align}$

$\begin{align}4x&=0\\x&=0\end{align}$ or $\begin{align}x-2&=0\\x&=2\end{align}$

Example

Solve $2x^3-14x^2+24x=0$

Show Solution

$\begin{align}2x^3-14x^2+24x&=0&&\text{Factor out the GCF of }2x\\2x(x^2-7x+12)&=0&&\text{Factor the quadratic expression in the parentheses}\\2x(x-3)(x-4)&=0&&\text{Set each factor equal to 0 and solve}\end{align}$

$\begin{align}2x=0\\x=0\end{align}$ or $\begin{align}x-3=0\\x=3\end{align}$ or $\begin{align}x-4=0\\x=4\end{align}$

Note that when we factor out and x (including higher powers of x), one of the solutions will be 0.

Example

Solve $6x^2+21x-27=0$

Show Solution

$\begin{align}6x^2+21x-27&=0&&\text{Factor out the GCF of 3}\\3(2x^2+7x-9)&=0&&\text{Factor the quadratic expression in the parentheses}\\3(2x+9)(x-1)&=0&&\text{3 cannot be equal to 0, so set the other factors equal to 0 and solve}\end{align}$

$\begin{align}2x+9=0\\x=-\frac{9}{2}\end{align}$ or $\begin{align}x-1=0\\x=1\end{align}$

In the previous example, the GCF did not have a variable in it, so no solutions come from this factor. You can skip setting the GCF factor equal to zero if there is no variables in the GCF.

Just as not all polynomials cannot factor, all equations cannot be solved by factoring. If an equation does not factor we will have to solve it using another method. These other methods are saved for another section.

World View Note: While factoring works great to solve problems with x², Tartaglia, in 16th century Italy, developed a method to solve problems with x³. 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.

Module 8: Polynomials and Polynomial Functions