Finding Polynomial Roots

Hello!

In our last lesson, we focused on applying our knowledge of linear and quadratic functions to model real-world problems. We saw how to translate a word problem into a mathematical equation, which is a crucial skill for problem-solving. While we were often given the quadratic function, a key part of solving those problems was finding the roots (or x-intercepts).

Today, we're going to build on that idea significantly. What if you have a polynomial of a higher degree, like a cubic (degree 3) or a quartic (degree 4)? How do you find its roots? Today's lesson will give you a powerful and systematic method to do just that.

Our learning outcome is to find the roots of polynomial functions using factorising (factoring) and the Rational Root Theorem. This method combines the Factor Theorem and synthetic division you've seen before with a new tool that cleverly narrows down the possibilities, making an otherwise impossible guessing game quite manageable.

1. The Challenge: Finding Roots of Higher-Degree Polynomials

You are already comfortable finding the roots of a quadratic equation like $x^{2} - 5 x + 6 = 0$ by factorising it into $(x - 2) (x - 3) = 0$ , which tells us the roots are $x = 2$ and $x = 3$ .

But what about a cubic polynomial like this one?
$P (x) = x^{3} + 2 x^{2} - 5 x - 6$
How would you find the values of $x$ that make $P (x) = 0$ ? Simple factorising by grouping doesn't work here. We could try guessing values for $x$ , but we might be guessing for a very long time. We need a more structured approach. This is where the Rational Root Theorem comes in.

2. The Rational Root Theorem: A Guide for Our Guesses

The Rational Root Theorem provides a complete list of all possible rational roots for a polynomial with integer coefficients. Instead of guessing from an infinite pool of numbers, it gives us a finite, and often small, list to test.

The theorem states that if a polynomial has a rational root, it must be of the form $\frac{p}{q}$ , where:

$p$ is a factor of the constant term.
$q$ is a factor of the leading coefficient (the coefficient of the term with the highest power of $x$ ).

Let's look at a visual representation of this.

Rational Root Theorem Formula — This diagram shows the general form of a polynomial, highlighting the constant term ($a_0$) and the leading coefficient ($a_n$). It illustrates that the possible rational roots are fractions formed by the factors of these two terms.

Let's apply this to our example polynomial: $P (x) = 1 x^{3} + 2 x^{2} - 5 x - 6$ .

Constant term: -6. Its factors ( $p$ ) are $\pm 1, \pm 2, \pm 3, \pm 6$ .
Leading coefficient: 1. Its factors ( $q$ ) are $\pm 1$ .

The possible rational roots ( $\frac{p}{q}$ ) are therefore:
$\frac{{ \pm 1 , \pm 2 , \pm 3 , \pm 6 }}{{ \pm 1 }} = {\pm 1, \pm 2, \pm 3, \pm 6}$
Instead of infinite possibilities, we now have just eight numbers to test!

For a more formal definition and another example, please read the following section from the Sparkl.me resource.

Polynomial division, factor theorem and remainder theorem

The resource 'Polynomial division, factor theorem and remainder theorem' offers a clear definition of the Rational Root Theorem. This will formalise what we've just discussed.

Please read section 8, 'The Rational Root Theorem', under the 'Advanced Concepts' heading. Focus on the definition and the example provided for f(x) = 2x^3 - 3x^2 - 8x + 12. Notice how the list of possible roots includes fractions because the leading coefficient is not 1.

3. The Full Algorithm: Find, Divide, and Factor

Now that we have a list of possible roots, we can use a step-by-step process, almost like an algorithm, to find all the roots of the polynomial.

List: Use the Rational Root Theorem to list all possible rational roots.
Test: Use the Factor Theorem to test the values from your list. (Remember, if $P (c) = 0$ , then $c$ is a root and $(x - c)$ is a factor).
Divide: Once you find a root $c$ , use synthetic division to divide the polynomial by the factor $(x - c)$ . The remainder should be 0.
Repeat/Factor: You'll be left with a polynomial of a lower degree (the quotient). If it's a quadratic, you can find the remaining roots by factorising or using the quadratic formula. If it's still a higher-degree polynomial, you repeat the process from Step 2 with the new, simpler polynomial.

This entire workflow is demonstrated very clearly in the following video.

Rational Root Theorem

The video 'Rational Root Theorem' from Mario's Math Tutoring provides a complete walkthrough of this process on a single example. It's a great way to see all the steps come together.

Please watch the entire video (about 5 minutes). Pay attention to how he: Lists the possible rational roots (which he calls 'zeros'). Uses synthetic division to test a possible root (he finds that x=1 works). Takes the resulting quadratic and factors it to find the other two roots. Finally, writes the polynomial in its fully factorised form.

4. A Complete Walkthrough

Let's consolidate this by working through another example together. Let's find the roots of $P (x) = x^{3} - 4 x^{2} - 3 x + 18$ .

Step 1: List the possible rational roots.

Factors of the constant term (18): $p = {\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18}$
Factors of the leading coefficient (1): $q = {\pm 1}$
Possible roots $\frac{p}{q}$ : ${\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18}$

Step 2: Test the possible roots.
Let's try a few values.

$P (1) = (1)^{3} - 4 (1)^{2} - 3 (1) + 18 = 1 - 4 - 3 + 18 = 12 \neq = 0$ . So, 1 is not a root.
$P (- 1) = (- 1)^{3} - 4 (- 1)^{2} - 3 (- 1) + 18 = - 1 - 4 + 3 + 18 = 16 \neq = 0$ . So, -1 is not a root.
$P (2) = (2)^{3} - 4 (2)^{2} - 3 (2) + 18 = 8 - 16 - 6 + 18 = 4 \neq = 0$ . So, 2 is not a root.
$P (- 2) = (- 2)^{3} - 4 (- 2)^{2} - 3 (- 2) + 18 = - 8 - 16 + 6 + 18 = 0$ . Success! -2 is a root.

Step 3: Divide the polynomial.
Since $x = - 2$ is a root, $(x + 2)$ is a factor. We use synthetic division to divide $x^{3} - 4 x^{2} - 3 x + 18$ by $(x + 2)$ .

-2 | 1  -4   -3   18
   |    -2   12  -18
   ------------------
     1  -6    9    0

The remainder is 0, as expected. The quotient is $x^{2} - 6 x + 9$ .

Step 4: Factor the remaining quadratic.
We now have $P (x) = (x + 2) (x^{2} - 6 x + 9)$ .
We can factor the quadratic: $x^{2} - 6 x + 9 = (x - 3) (x - 3) = (x - 3)^{2}$ .

So, the fully factorised form is $P (x) = (x + 2) (x - 3)^{2}$ .
The roots are -2 and 3 (where 3 is a "repeated root" or a root with multiplicity 2, a concept we will explore more later).

The following image shows another complete example, this time for a 4th-degree (quartic) polynomial. Notice how synthetic division is used twice to get down to a quadratic.

Factoring a Polynomial using Rational Root Theorem and Synthetic Division — This image shows the full process for factoring $f(x) = x^4 - x^3 - 12x^2 - 4x + 16$. It starts by listing possible roots, then uses synthetic division with $x=1$ and $x=-2$ to reduce the polynomial to the quadratic $x^2 - 4$, which is then factored to find the remaining roots.

Test your understanding!

Now it's your turn to apply the full algorithm. Find all the roots of the polynomial:
$P (x) = 2 x^{3} + x^{2} - 13 x + 6$
Hint: Remember that the leading coefficient is 2, so you will have some fractional possible roots to consider.

Show answer

Step 1: List possible rational roots.

Factors of constant term (6): $p = {\pm 1, \pm 2, \pm 3, \pm 6}$
Factors of leading coefficient (2): $q = {\pm 1, \pm 2}$
Possible roots $\frac{p}{q}$ : ${\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}}$

Step 2: Test the possible roots.
Let's try $x = 2$ :
$P (2) = 2 (2)^{3} + (2)^{2} - 13 (2) + 6 = 2 (8) + 4 - 26 + 6 = 16 + 4 - 26 + 6 = 0$ .
Great! We found a root: 2.

Step 3: Divide the polynomial.
We use synthetic division to divide by $(x - 2)$ .

2 | 2   1   -13   6
  |     4    10  -6
  -----------------
    2   5    -3   0

The quotient is $2 x^{2} + 5 x - 3$ .

Step 4: Factor the remaining quadratic.
We need to solve $2 x^{2} + 5 x - 3 = 0$ . We can factor this. We look for two numbers that multiply to $2 \times - 3 = - 6$ and add to 5. These numbers are 6 and -1.
We rewrite the middle term:
$2 x^{2} + 6 x - x - 3 = 0$
Factor by grouping:
$2 x (x + 3) - 1 (x + 3) = 0$
$(2 x - 1) (x + 3) = 0$
This gives us two more roots: $x = \frac{1}{2}$ and $x = - 3$ .

The roots of the polynomial are 2, -3, and $\frac{1}{2}$ .

Conclusion

Today we've learned a systematic method for finding the rational roots of any polynomial with integer coefficients. This "algorithm" is a huge step up from simply guessing and is a cornerstone of analysing polynomial functions.

Key Takeaways:

The Rational Root Theorem dramatically narrows down the search for rational roots by creating a list of all possible candidates based on the factors of the constant term and the leading coefficient.
The complete process for finding roots is to List (using the theorem), Test (using the Factor Theorem), Divide (using synthetic division), and Factor/Repeat.
This process works for any degree of polynomial, allowing you to break down complex cubics or quartics into simpler quadratics that you already know how to solve.

Preview of the Next Lesson:
Finding the roots gives us the crucial x-intercepts of a polynomial's graph. But what does the graph do between and beyond these roots? In our next lesson, we will analyse the end behaviour and turning points of polynomial functions. This will allow us to start sketching the overall shape of these more complex curves.

Recommended resources

1. The Challenge: Finding Roots of Higher-Degree Polynomials

2. The Rational Root Theorem: A Guide for Our Guesses

3. The Full Algorithm: Find, Divide, and Factor

4. A Complete Walkthrough

Conclusion