Rational Roots Theorem

Consider the polynomial equation below where all coefficients The_noare integers:

The_nox^no + the_n-1x^n-1 + the_n-2x^n-2 + … + the₂x² + the₁x + a₀ = 0

O Rational Roots Theorem guarantees that if this equation admits the rational number ^P/_what as root (with P, what  and mdc (p, q) = 1), then The₀ is divisible by P and The_no is divisible by what.

Comments:

1º) The rational roots theorem does not guarantee that the polynomial equation has roots, but if they do exist, the theorem allows us to identify all roots of the equation;

2º) if The_no= 1 and the other coefficients are all integers, the equation has only integer roots.

3°) if q = 1 and there are rational roots, these are whole and divisors of The₀.

Application of the Rational Roots Theorem:

Let's use the theorem to find all the roots of the polynomial equation 2x⁴ + 5x³ – 11x² – 20x + 12 = 0.

First, let's identify the possible rational roots of this equation, that is, the roots of the form ^P/_what. According to the theorem, The₀ is divisible by P; in this way, how

The₀ = 12, then the possible values ​​of P are {±1, ±2, ±3, ±4, ±6, ±12}. Analogously, we have to The_no is divisible by what and The_no = 2, then what can have the following values: {±1, ±2}. Therefore, dividing the values ​​of P per what, we get possible values ^P/_what roots of the equation: {+½, – ½, +1, – 1, +³/_{2, –}³/₂, +2, –2, +3, –3, +4, –4, +6, –6, +12, –12}.

To confirm that the values ​​we found are really the root of the polynomial equation, let's substitute each value in place of the x of the equation. Through algebraic calculus, if the polynomial results in zero, so the substituted number is actually the root of the equation.

2x⁴ + 5x³ – 11x² – 20x + 12 = 0

For x = + ½

2.(½)⁴ + 5.(½)³ – 11.(½)² – 20.(½) + 12 = 0

For x = – ½

2.(– ½)⁴ + 5.(– ½)³ – 11.(– ½)² – 20.(– ½) + 12 = ⁷⁵/₄

For x = + 1

2.1⁴ + 5.1³ – 11.1² – 20.1 + 12 = – 12

For x = – 1

2.(– 1)⁴ + 5.(– 1)³ – 11.(– 1)² – 20.(– 1) + 12 = 18

For x = + ³/₂

2.(³/₂)⁴ + 5.(³/₂)³ – 11.(³/₂)² – 20.(³/₂) + 12 = – ⁶³/₄

For x = - ³/₂

2.(– ³/₂)⁴ + 5.(– ³/₂)³ – 11.(– ³/₂)² – 20.(– ³/₂) + 12 = ²¹/₂

For x = + 2

2.2⁴ + 5.2³ – 11.2² – 20.2 + 12 = 0

For x = – 2

2.(– 2)⁴ + 5.(– 2)³ – 11.(– 2)² – 20.(– 2) + 12 = 0

For x = + 3

2.3⁴ + 5.3³ – 11.3² – 20.3 + 12 = 150

For x = – 3

2.(– 3)⁴ + 5.(– 3)³ – 11.(– 3)² – 20.(– 3) + 12 = 0

For x = + 4

2.4⁴ + 5.4³ – 11.4² – 20.4 + 12 = 588

For x = – 4

2.(– 4)⁴ + 5.(– 4)³ – 11.(– 4)² – 20.(– 4) + 12 = 108

For x = + 6

2.6⁴ + 5.6³ – 11.6² – 20.6 + 12 = 3168

For x = – 6

2.(– 6)⁴ + 5.(– 6)³ – 11.(– 6)² – 20.(– 6) + 12 = 1248

For x = + 12

2.12⁴ + 5.12³ – 11.12² – 20.12 + 12 = 48300

For x = – 12

2.(– 12)⁴ + 5.(– 12)³ – 11.(– 12)² – 20.(– 12) + 12 = 31500

Therefore, the roots of the polynomial equation 2x⁴ + 5x³ – 11x² – 20x + 12 = 0 they are {– 3, – 2, ½, 2}. Through polynomial decomposition theorem, we could write this equation as (x + 3).(x + 2).(x – ½).(x – 2)= 0.

By Amanda Gonçalves
Graduated in Mathematics

Source: Brazil School - https://brasilescola.uol.com.br/matematica/teorema-das-raizes-racionais.htm