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Math Lesson 11.3.2 - The Rational Zero (Root) Theorem

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Welcome to our Math lesson on The Rational Zero (Root) Theorem, this is the second lesson of our suite of math lessons covering the topic of Solutions for Polynomial Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

The Rational Zero (Root) Theorem

The Rational Zero Theorem (otherwise known as the Rational Root Theorem), as its name suggests, is used to find rational solutions when dealing with a polynomial equation (or zeros or roots of a polynomial expression).

A polynomial does not always have zeroes. This means that the corresponding polynomial equations do not always have zeroes. But when they have, these zeroes must be rational numbers. Given that the general form of a polynomial is

P(x) = an xn + an - 1 xn - 1 + ⋯ + a1 x1 + a0 x0

or

P(x) = an xn + an - 1 xn - 1 + ⋯ + a1 x + a0

From the previous tutorials, it is known that all values of the variable x for which P(x) = 0 are the zeroes (roots) of the polynomial. For example, the values -3, -1 and 1 are the zeroes of the polynomial

P(x) = -x3 - 3x2 + x + 3

because

P(-3) = -(-3)3 - 3(-3)2 + (-3) + 3
= -(-27) - 3 ∙ 9 + (-3) + 3
= 27 - 27 - 3 + 3
= 0
P(-1) = -(-1)3-3(-1)2 + (-1) + 3
= -(-1) - 3 ∙ 1 + (-1) + 3
= 1 - 3 - 1 + 3
= 0

and

P(1) = -13-3 ∙ 12 + 1 + 3
= -1 - 3 ∙ 1 + 1 + 3
= -1 - 3 + 1 + 3
= 0

The Rational Root Theorem states that all rational roots of a polynomial equation with integer coefficients have the form p/q, where p is a factor of the polynomial constant a0 and q is a factor of the leading coefficient an. The two numbers p and q must be relatively prime.

Proof

First, let's prove that p is a factor of the constant a0. Indeed, substituting the variable x with the assumed root p/q yields

P(x) = an (p/q)n + an - 1 (p/q)n - 1 + ⋯ + a1 (p/q) + a0 = 0

Multiplying both sides by qn yields

P(p/q) = an (p/q)n ∙ qn + an - 1 (p/q)n - 1 ∙ qn + ⋯ + a1 (p/q) ∙ q1 + a0 ∙ q0 = 0
P(p/q) = an ∙ pn + an - 1 ∙ pn - 1 ∙ q1 + ⋯ + a1 ∙ p1 ∙ qn - 1 + a0 ∙ qn = 0

or

an ∙ pn + an - 1 ∙ pn - 1 ∙ q1 + ⋯ + a1 ∙ p1 ∙ qn - 1 = -a0 ∙ qn

Since p divides each term on the left side, it is a factor of the left side of the above expression. This must be true for the right side as well since both sides are equivalent. This means p is also a factor of the term (-a0 · qn). Hence, since p and q are relatively prime and cannot divide each other, the only possibility is that p is a factor of the constant a0.

Now, let's prove that q is a factor of the leading coefficient an. For this, we subtract an · pn from both sides of the polynomial

an ∙ pn + an - 1 ∙ pn - 1 ∙ q1 + ⋯ + a1 ∙ p1 ∙ qn - 1 + a0 ∙ qn = 0

This yields,

an - 1 ∙ pn - 1 ∙ q1 + ⋯ + a1 ∙ p1 ∙ qn - 1 + a0 ∙ qn = -an ∙ pn

Since q is a factor of all terms on the left side, it is also a factor of the entire expression written on the left. Therefore, it must be a factor of the right side as well, since both sides are equivalent. Given that p and q are relatively prime (they cannot have common factors), then q is a factor of the leading coefficient an.

For example, in the polynomial

P(x) = -x3 - 3x2 + x + 3

considered earlier in this tutorial, we have an = -1 and a0 = 3 (n = 3, as it represents the degree of the polynomial). Hence, all roots must be multiples of -1 or 3. Indeed, we proved by substitution that -3, -1 and 1 are roots for the given polynomial.

More Solutions for Polynomial Equations Lessons and Learning Resources

Polynomials Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
11.3Solutions for Polynomial Equations
Lesson IDMath Lesson TitleLessonVideo
Lesson
11.3.1The Synthetic Division Method of Polynomials
11.3.2The Rational Zero (Root) Theorem
11.3.3How to list all Possible Zeroes (Roots) of a Polynomial Equation
11.3.4How to identify the Zeroes of the Polynomial from the List of Possible Zeroes
11.3.5How to Solve Polynomial Equations

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