Math Calculators ▶ Rational Zeros Calculator
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The rational zeros calculator finds all possible rational roots of a polynomial and lets you know which of these are actual. For the polynomial you enter, the tool will apply the rational zeros theorem to validate the actual roots among all possible values.
A rational zero is a number in the form of p/q which on putting in the original polynomial yields zero.
$$ P\left(x\right) = a_{n}x^{n} + a_{n-1}x^{n-1} + … + a_{1}x + a_{0} \hspace{0.25in}where\hspace{0.25in} \left(a_{n}≠0\right) $$
The standard rational root theorem satisfies the following conditions:
Let us resolve an example that will help you calculate all possible and actual roots of the function given:
\(2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
Factors of -2 = \(+1, -1, +2, -2\) (These factors are values of “p”)
Factors of 2 = \(+1, -1, +2, -2\) (These factors are values of “q”)
\(\dfrac{1}{-1}, \dfrac{-1}{-1}, \dfrac{2}{-1}, \dfrac{-2}{-1}, \dfrac{1}{2}, \dfrac{-1}{2}, \dfrac{2}{2}, \dfrac{-2}{2}, \dfrac{1}{-2}, \dfrac{-1}{-2}, \dfrac{2}{-2}, \dfrac{-2}{-2}\)
\(\dfrac{1}{-1}, \dfrac{-1}{-1}, \dfrac{2}{-1}, \dfrac{-2}{-1}, \dfrac{1}{2}, \dfrac{-1}{2}, \dfrac{2}{2}, \dfrac{-2}{2}, \dfrac{1}{-2}, \dfrac{-1}{-2}, \dfrac{2}{-2}, \dfrac{-2}{-2}\)
\(\dfrac{1}{2}, 1, 2, -2, -1, \dfrac{-1}{2}\)
\(Root \dfrac{1}{2}:\)
\(P\left(\drac{1}{2}\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(\drac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)^{6}+7\left(\dfrac{1}{2}\right)^{5}+\left(\dfrac{1}{2}\right)^{4}-3\left(\dfrac{1}{2}\right)^{3}+6\left(\dfrac{1}{2}\right)^{2}+2\left(\dfrac{1}{2}\right)-2\)
\(P\left(\drac{1}{2}\right) = 1.25\)
\(P\left(1\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(1\right) = 2\left(1\right)^{6}+7\left(1\right)^{5}+\left(1\right)^{4}-3\left(1\right)^{3}+6\left(1\right)^{2}+2\left(1\right)-2\)
\(P\left(1\right) = 13\)
\(P\left(2\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(2\right) = 2\left(2\right)^{6}+7\left(2\right)^{5}+\left(2\right)^{4}-3\left(2\right)^{3}+6\left(2\right)^{2}+2\left(2\right)-2\)
\(P\left(2\right) = 370\)
\(P\left(-2\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(-2\right) = 2\left(-2\right)^{6}+7\left(-2\right)^{5}+\left(-2\right)^{4}-3\left(-2\right)^{3}+6\left(-2\right)\left(-2\right)^{2}+2\left(-2\right)-2\)
\(P\left(-2\right) = -370\)
\(P\left(-1\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(-1\right) = 2\left(-1\right)^{6}+7\left(-1\right)^{5}+\left(-1\right)^{4}-3\left(-1\right)^{3}+6\left(-1\right)^{2}+2\left(-1\right)-2\)
\(P\left(1\right) = -13\)
\(P\left(\drac{-1}{2}\right) = 2x^{6}+7x^{5}+x^{4}-3x^{3}+6x^{2}+2x-2\)
\(P\left(\drac{-1}{2}\right) = 2\left(\dfrac{-1}{2}\right)^{6}+7\left(\dfrac{-1}{2}\right)^{5}+\left(\dfrac{-1}{2}\right)^{4}-3\left(\dfrac{-1}{2}\right)^{3}+6\left(\dfrac{-1}{2}\right)^{2}+2\left(\dfrac{-1}{2}\right)-2\)
\(P\left(\drac{1}{2}\right) = -1.25\)
Hence proved that there exist no actual roots that fully satisfy the given polynomial. You can also put in the given statement in the rational zero theorem calculator to verify your calculations.
Our possible zeros calculator functions to display instant and precise calculations for rational zeros. Yes, it is possible if you follow a couple of steps:
Input:
Output:
A rational zero is one which has terminating decimal places in it. On the other hand, an irrational zero has non-terminating decimal places in it. With this rational zeros theorem calculator, you can not only determine all possible roots but can get a clear difference between rational and irrational roots.
From the source Wikipedia: Rational root
From the source Khan Academy: Zeros of polynomials: matching equation to graph, Zeros with factoring
From the source Lumen Learning: Polynomial Function, The Fundamental Theorem of Algebra