Question

What are the roots of the polynomial equation x3 - 6x = 3x2 - 8? Use a graphing calculator and a system of equations.

a. –40, –4, 5
b. –5, 4, 40
c. –4, –1, 2
d. –2, 1, 4

Answer 1

d : 4, 1, -2.

Explanation:

There's no need for a calculator in my opinion because we can use the rational root theorem which states that the equations of this form:

`\sum_(i=0)^(n) a_(i)x^i = 0 = a_(0) + a_(1)x + a_2x^2 + ... + a_nx^n = 0\\ a_i \in \mathbb{Z}`

Have rational roots, than the roots are of the form:
`(k \cdot a_o)/(p \cdot a_n), \ where \ k, p \in \mathbb{Z}.`

Rewriting the equation we have:

`x^3 - 3x^2 -6x + 8 = 0\\By \ our\ previous \ claim \ x \in D_8 \ where \ D_n \ the \ set \ of \ divisors \ of \ n.\\D_8 = \{\pm 1, \pm 2, \pm 4, \pm 8\} \\We \ plug \ in \ some \ number from \ D_8.\\Letting \ x = -2;\\(-2)^3 -3(-2)^2 -6(-2) + 8 = 0 \ \ \ Thus \ x+2 \ is \ a \ factor.\\We \ can \ now \ simplify \ the \ eq. \ using \ Polynomial \ Long \ Division.`

`x^3 -3x^2 -6x + 8 = (x+2)(Q(x))\\To \ find \ Q \ we \ divide \ the \ original \ equation \ by \ x + 2; which \ yields:\\Q(x) = x^2 - 5x + 4.\\So \ we \ can \ use \ the \ quadratic \ formula \ to \ find \ the \ roots:\\x = 4,x = 1. \\Thus \ x \in \{-2, 1, 4\}. \ \ These \ are \ all \ the \ roots.`