Answer:
(a) [tex]x\³ - 6x - 6[/tex]
(b) Proved
Step-by-step explanation:
Given
[tex]r = $\sqrt[3]{2} + \sqrt[3]{4}$[/tex] --- the root
Solving (a): The polynomial
A cubic function is represented as:
[tex]f = (a + b)^3[/tex]
Expand
[tex]f = a^3 + 3a^2b + 3ab^2 + b^3[/tex]
Rewrite as:
[tex]f = a^3 + 3ab(a + b) + b^3[/tex]
The root is represented as:
[tex]r=a+b[/tex]
By comparison:
[tex]a = $\sqrt[3]{2}[/tex]
[tex]b = \sqrt[3]{4}$[/tex]
So, we have:
[tex]f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3[/tex]
Expand
[tex]f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4[/tex]
[tex]f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4[/tex]
[tex]f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4[/tex]
[tex]f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4[/tex]
Evaluate like terms
[tex]f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)[/tex]
Recall that: [tex]r = $\sqrt[3]{2} + \sqrt[3]{4}$[/tex]
So, we have:
[tex]f = 6 + 6r[/tex]
Equate to 0
[tex]f - 6 - 6r = 0[/tex]
Rewrite as:
[tex]f - 6r - 6 = 0[/tex]
Express as a cubic function
[tex]x^3 - 6x - 6 = 0[/tex]
Hence, the cubic polynomial is:
[tex]f(x) = x^3 - 6x - 6[/tex]
Solving (b): Prove that r is irrational
The constant term of [tex]x^3 - 6x - 6 = 0[/tex] is -6
The divisors of -6 are: -6,-3,-2,-1,1,2,3,6
Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values
[tex]f(-6) = (-6)^3 - 6*-6 - 6 = -186[/tex]
[tex]f(-3) = (-3)^3 - 6*-3 - 6 = -15[/tex]
[tex]f(-2) = (-2)^3 - 6*-2 - 6 = -2[/tex]
[tex]f(-1) = (-1)^3 - 6*-1 - 6 = -1[/tex]
[tex]f(1) = (1)^3 - 6*1 - 6 = -11[/tex]
[tex]f(2) = (2)^3 - 6*2 - 6 = -10[/tex]
[tex]f(3) = (3)^3 - 6*3 - 6 = 3[/tex]
[tex]f(6) = (6)^3 - 6*6 - 6 = 174[/tex]
For r to be rational;
The divisors of -6 must divide f(x) without remainder
i.e. Any of the above values must equal 0
Since none equals 0, then r is irrational
