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Solving Polynomials

Q: Can a polynomial have no real solutions?

Yes. For example, the equation $x^2 + 9 = 0$ has no real number solutions. This is because you can't square a real number and get a negative result, but its solutions do exist in the complex number system.

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Ever looked at an equation like $x^{2} + 5 x + 6 = 0$ and wondered where to even begin? You're in the right place! This guide will unlock the secrets of solving polynomial equations, turning complex problems into simple, step-by-step puzzles you can confidently solve.

Solving Polynomials — an original Algebra911 reference diagram defining solving polynomials with its key formula and a worked example. — Solving Polynomial Equations: An Algebra 911 Guide

What Is a Polynomial Equation and What Does It Mean to 'Solve' It?

Solving a polynomial equation means finding the specific numerical values of the variable that make the equation a true statement. These values are called the solutions or roots of the equation. They are also sometimes called the zeros of the polynomial, because they are the values that make the polynomial expression equal to zero.

First, let's clarify the difference between an expression and an equation. A polynomial expression is a collection of terms with variables and coefficients, like $5 x^{3} - x^{2} + 8 x - 4$ . It doesn't have an equals sign. A polynomial equation, on the other hand, sets the polynomial equal to something, most often zero, like $5 x^{3} - x^{2} + 8 x - 4 = 0$ . Our goal is to find the value(s) of $x$ that work.

The degree of a polynomial is the highest exponent on the variable. This number is important because it tells you the maximum number of solutions the equation can have. For example:

$3 x - 12 = 0$ is a linear equation (degree $1$ ) and has one solution.
$x^{2} + 5 x + 6 = 0$ is a quadratic equation (degree $2$ ) and can have up to two solutions.
$x^{3} - 8 = 0$ is a cubic equation (degree $3$ ) and can have up to three solutions.

To solve these equations, we will use a set of powerful algebraic tools. Let's start with the most fundamental rule of them all.

The Zero Product Property: Your Most Important Tool

The single most important concept for solving polynomial equations is the Zero Product Property. It's a simple idea with powerful consequences. It states that if the product of two or more factors is zero, then at least one of those factors must be zero.

In mathematical terms, if $A \cdot B = 0$ , then either $A = 0$ or $B = 0$ (or both are zero). Think about it: you can't multiply two non-zero numbers together and get zero as the answer. This property is our key to unlocking the solutions hidden inside a factored polynomial.

For example, imagine you're given the equation already factored like this:

$(x - 5) (x + 2) = 0$

Here, our $A$ is $(x - 5)$ and our $B$ is $(x + 2)$ . According to the Zero Product Property, for the whole thing to equal zero, one of those pieces must be zero. So, we set up two separate, smaller equations:

$x - 5 = 0$ OR $x + 2 = 0$

Solving each one is simple:

$x = 5$ OR $x = - 2$

And just like that, we found the two solutions! The values $x = 5$ and $x = - 2$ are the roots of the original equation. This is why our first major strategy involves getting our polynomial into a factored form set equal to zero.

How Do You Solve Polynomials by Factoring?

Factoring is the process of breaking a polynomial down into a product of simpler expressions (its factors). If we can factor a polynomial and set it equal to zero, we can use the Zero Product Property to find the solutions. This is often the quickest and easiest method.

Here is the general process:

Set the Equation to Zero: Make sure one side of the equation is $0$ . If it's not, use addition or subtraction to move all terms to one side.
Factor Completely: Use your factoring skills to break down the polynomial. Always start by looking for a Greatest Common Factor (GCF). Then, depending on the number of terms, use other techniques like:
- Difference of Squares: For expressions like $a^{2} - b^{2}$ , which factors to $(a - b) (a + b)$ .
- Trinomial Factoring: For expressions like $x^{2} + b x + c$ , find two numbers that multiply to $c$ and add to $b$ .
Apply the Zero Product Property: Set each individual factor equal to zero.
Solve for the Variable: Solve each of the new, simpler equations to find all the solutions.

Example 1

Solve the equation $x^{2} - 3 x = 10$ .

Step 1: Set the equation to zero.

We need to move the $10$ to the left side by subtracting it from both sides.

$x^{2} - 3 x - 10 = 0$

Step 2: Factor the polynomial.

This is a trinomial. We need to find two numbers that multiply to $- 10$ and add to $- 3$ . Let's think about the factors of $- 10$ : $(1, - 10)$ , $(- 1, 10)$ , $(2, - 5)$ , and $(- 2, 5)$ . The pair that adds up to $- 3$ is $2$ and $- 5$ .

$(x + 2) (x - 5) = 0$

Step 3: Apply the Zero Product Property.

Now we set each factor equal to zero.

$x + 2 = 0 or x - 5 = 0$

Step 4: Solve for $x$ .

Solving the two simple equations gives us our solutions.

$x = - 2 or x = 5$

The two solutions to the equation $x^{2} - 3 x = 10$ are $- 2$ and $5$ .

What if Factoring Fails? The Quadratic Formula to the Rescue!

Sometimes you'll encounter a quadratic equation that is difficult or impossible to factor using simple integers. Does that mean it has no solution? Not at all! For any quadratic equation written in the standard form $a x^{2} + b x + c = 0$ , we have a powerful tool that works every single time: the Quadratic Formula.

This formula can look intimidating at first, but it's just a matter of carefully plugging in numbers. For an equation in the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are the coefficients, the solutions for $x$ are given by:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The $\pm$ symbol means we have to perform the calculation twice: once with a plus sign and once with a minus sign. This is how we often get two distinct solutions.

Example 2

Solve the equation $3 x^{2} - 5 x - 1 = 0$ .

Step 1: Identify $a$ , $b$ , and $c$ .

The equation is already in standard form, $a x^{2} + b x + c = 0$ . By comparing, we can see:

$a = 3$
$b = - 5$
$c = - 1$

Step 2: Substitute these values into the Quadratic Formula.

Be very careful with the signs, especially with $- b$ when $b$ is already negative.

$x = \frac{- (- 5) \pm \sqrt{(- 5)^{2} - 4 (3) (- 1)}}{2 (3)}$

Step 3: Simplify the expression.

Work through the arithmetic step by step, following the order of operations.

$x = \frac{5 \pm \sqrt{25 - (- 12)}}{6}$

$x = \frac{5 \pm \sqrt{25 + 12}}{6}$

$x = \frac{5 \pm \sqrt{37}}{6}$

Step 4: State the final solutions.

Since $37$ is a prime number, its square root cannot be simplified further. So, we write the two exact solutions separately.

$x = \frac{5 + \sqrt{37}}{6} and x = \frac{5 - \sqrt{37}}{6}$

These are the two real, irrational solutions to the equation. Using a calculator, they are approximately $x \approx 1.847$ and $x \approx - 0.180$ .

Can You Solve Polynomials with a Degree Higher Than 2?

Yes! The same fundamental principles apply. The main goal is always to break the polynomial down into a product of linear or quadratic factors and then use the Zero Product Property. For cubic (degree 3) or quartic (degree 4) polynomials, factoring can be more complex, but there are specific techniques that often work.

One common method for polynomials with four terms is Factoring by Grouping. Here's how it works:

Arrange the terms in descending order of their exponents.
Group the first two terms together and the last two terms together.
Factor out the GCF from each group.
If the binomials inside the parentheses match, you can factor that common binomial out, leaving the GCFs as the other factor.

Let's see this powerful technique in action.

Example 3

Solve the cubic equation $x^{3} + 4 x^{2} - 9 x - 36 = 0$ .

Step 1: Group the terms.

The equation is already set to zero and the terms are in order. We'll group them into two pairs.

$(x^{3} + 4 x^{2}) + (- 9 x - 36) = 0$

Step 2: Factor the GCF from each group.

In the first group, the GCF is $x^{2}$ . In the second group, the GCF is $- 9$ . Factoring these out gives us:

$x^{2} (x + 4) - 9 (x + 4) = 0$

Step 3: Factor out the common binomial.

Notice that both parts now share the factor $(x + 4)$ . We can factor this out.

$(x + 4) (x^{2} - 9) = 0$

Step 4: Factor completely.

We're not done yet! The second factor, $x^{2} - 9$ , is a difference of squares ( $x^{2} - 3^{2}$ ). It can be factored further.

$(x + 4) (x - 3) (x + 3) = 0$

Step 5: Apply the Zero Product Property.

Now that we have three linear factors, we set each one equal to zero.

$x + 4 = 0 or x - 3 = 0 or x + 3 = 0$

Step 6: Solve.

Solving each simple equation gives us our three solutions.

$x = - 4, x = 3, x = - 3$

The cubic equation has three solutions, just as we expected from its degree.

What Are Some Common Mistakes When Solving Polynomials?

Solving polynomials requires careful attention to detail. A small error can lead to a completely wrong answer. Here are some of the most common pitfalls to watch out for:

Forgetting to Set the Equation to Zero: The Zero Product Property only works when the product is equal to $0$ . If you have $(x - 2) (x - 3) = 6$ , you cannot say $x - 2 = 6$ or $x - 3 = 6$ . You must first expand the left side, subtract $6$ from both sides to get $x^{2} - 5 x = 0$ , and then solve from there.
Sign Errors in the Quadratic Formula: This is the most frequent mistake. Be extremely careful with the $- b$ part and when calculating the discriminant ( $b^{2} - 4 a c$ ). If $b$ is negative, $- b$ becomes positive. If $a$ or $c$ is negative, the $- 4 a c$ term might become positive.
Incomplete Factoring: Always check if your factors can be factored further. Forgetting to factor a difference of squares, like in the $x^{2} - 9$ example above, will cause you to miss solutions. Also, always look for a GCF first—it makes the remaining polynomial much easier to handle.
Square Root Errors: When solving an equation like $x^{2} = 16$ by taking the square root, remember that there are always two solutions: $x = 4$ and $x = - 4$ . It's easy to forget the negative solution.
Distribution Errors: When expanding polynomials or checking your work, be careful to distribute terms correctly, especially negative signs.

Your Polynomial Toolkit: A Quick Summary

Navigating different types of polynomial equations requires choosing the right strategy. Here is a quick-reference table to summarize the methods we've discussed and when to use them.

Method	When to Use It	Key Idea
Factoring (GCF)	Always try this first, for any polynomial with a common factor in all terms.	Simplify the problem by pulling out the greatest common factor.
Trinomial / Difference of Squares Factoring	For quadratic equations (degree 2) that are in a recognizable pattern.	Break the polynomial into a product of two binomials.
Quadratic Formula	For any quadratic equation ( $a x^{2} + b x + c = 0$ ), especially when it doesn't factor easily.	A universal formula that solves any quadratic equation.
Factoring by Grouping	For some polynomials with 4 terms.	Group terms into pairs, find the GCF of each pair, and factor out a common binomial.

Frequently Asked Questions

What's the difference between a polynomial expression and a polynomial equation?

A polynomial expression is a mathematical phrase without an equals sign, like $x^{2} + 3 x$ . An equation sets two expressions equal to each other, like $x^{2} + 3 x = 4$ , and can be solved to find the values of the variable that make the statement true.

How many solutions can a polynomial equation have?

The Fundamental Theorem of Algebra states that a polynomial's total number of solutions (including real and complex numbers) is equal to its degree. For example, a cubic equation (degree 3) will have 3 solutions, while a quadratic (degree 2) has 2.

Can a polynomial have no real solutions?

Yes. For example, the equation $x^{2} + 9 = 0$ has no real number solutions. This is because you can't square a real number and get a negative result, but its solutions do exist in the complex number system.

Why must the equation be set to zero before factoring?

This step is crucial because it allows us to use the Zero Product Property. This property only works for a product equal to zero; if a product equals any other number, like 6, we cannot make any certain conclusions about the individual factors.

What is the 'degree' of a polynomial?

The degree is the highest exponent of the variable in any single term of the polynomial. For instance, in the polynomial $7 x^{4} - 2 x^{3} + 5$ , the highest exponent is 4, so the degree is 4.

Is there a formula for cubic equations like the quadratic formula?

Yes, a cubic formula does exist, but it is incredibly long and complex, so it's almost never used in high school algebra. Instead, we use methods like factoring by grouping or graphing to solve cubic equations.

What should I do if I can't factor a polynomial?

If it's a quadratic equation (degree 2), you should immediately use the quadratic formula, as it always works. For higher-degree polynomials that you can't factor, you may need more advanced techniques like the Rational Root Theorem or graphical methods, which are typically taught in more advanced algebra courses.