Rational Roots Test

The Rational Roots Test (also known as Rational Zeros Theorem) allows us to find all possible rational roots of a polynomial. Suppose $a$ is root of the polynomial $P\left( x \right)$ that means $P\left( a \right) = 0$ . In other words, if we substitute $a$ into the polynomial $P\left( x \right)$ and get zero, $0$ , it means that the input value is a root of the function.

But how do we find the possible list of rational roots? Here’s how it works in a nutshell!

Key Ideas of Rational Roots Test

Suppose we have some polynomial $P\left( x \right)$ with integer coefficients and a nonzero constant term:

in P(x), a sub n is the leading coefficient while a sub zero is a constant term

Then every rational root of $P\left( x \right)$ is of the form:

p/q = plus or minus factors of a sub zero divided by plus or minus factors of a sub n

The best way to learn this method is to take a look at some examples!

Examples of How to Find the Rational Roots of a Polynomial using the Rational Roots Test

Example 1: Find the rational roots of the polynomial below using the Rational Roots Test.

P(x) is equal to three x cubed minus four x squared minus seventeen x plus six

Finding the rational roots (also known as rational zeroes) of a polynomial is the same as finding the rational $x$ -intercepts.

Start by identifying the constant term ${a_0}$ and the leading coefficient ${a_n}$ .

in P(x), a sub n equals 3 , and a sub zero equals 6

Determine the positive and negative factors of each.

Factors of constant term, ${a_0} = 6\,\,:\,\, \pm \,\left( {1,2,3,6} \right)$

Factors of leading term, ${a_n} = 3\,\,:\,\, \pm \,\left( {1,3} \right)$

Write down the list of the possible rational roots by finding ${p \over q}$ which is simply the ratio of the factors of the constant term and leading term. Make sure that you keep track of the possible combinations.

This is how I do it. I take each numerator and divide it by all denominators. Then I move on to the next numerator and again divide by all denominators. I keep repeating this process until I have gone through all the numerators. This ensures that we have covered all possible combinations.

p/q = factors of 6 divided by factors of 3

BIG Caution: After you write down all combinations, simplify the fractions in order to get rid of duplicates.

get rid of the duplicate roots which in this situation are plus or minus 2 is the same as plus or minus 6/3

So these are the numbers without duplicates that we will check as possible roots. We have twelve (12) possible candidates to check.

all possible roots are 1/3, -1/3, 2/3, -2/3, 1, -1, 2, -2, 3, -3, 6, -6

Remember that if $a$ is a root of the polynomial $P\left( x \right)$ , then $P\left( a \right) = 0$ . Now, let’s check each number.

the roots of the polynomials are 1/3, -2, and 3 since P(1/3)=0, P(-2)=0, and P(3)=0.

Therefore, the rational roots of the polynomial

are

Here is the graph of the polynomial showing where it crosses or touches the $x$ -axis. These are in fact the $x$ -intercepts of the polynomial.

the polynomial P(x)=3x^3-4x^2-17x+6 crosses the x axis at -2, 1/3, 3

Example 2: Find the rational roots of the polynomial below using Rational Roots Test.

P(x) is equal to four x to the fifth power minus twelve to the fourth power plus 5 x cubed plus 8 x squared minus three x minus 2

The constant term is ${a_0} = - 2$ and its possible factors are $p = \pm 1,\, \pm 2$ . For the leading coefficient, we have ${a_n} = 4$ and its factors are $q = \pm 1,\, \pm 2,\, \pm 4$ .

To find the possible roots of the polynomial, write in the form

p/q = factors of -2 divided by factors of 4

Write down all possible

combinations:

Simplify each fraction to eliminate duplicates or identical values. Here’s our new and improved list!

p/q equals plus or minus 1/4, plus or minus 1/2, plus or minus 1, and plus or minus 2

Due to the plus or minus consideration of each number, we will have eight (8) possible candidates as the roots of this polynomial.

p/q are 1/4, -1/4, 1/2, -1/2, 1, -1, 2, -2

If you plug in each value to the given polynomial and get zero, that means the number you substituted is a root! Try this on paper, and you should be convinced that there are only three values satisfying this condition.

P(2)=4(2^5)-12(2^4)+5(2^3)+8(2^2)-3(2)-2=0

Therefore, the rational roots of the polynomial

are

Graphically, it shows that the polynomial touches or crosses the $x$ -axis at those roots determined by rational roots test.

the graph of P(x)=4x^5-12x^4+5x^3+8x^2-3x-2 has x intercepts of -1/2, 1, and 2