polynomials question (1 Viewer)

[nanashi]

hey i need help with a polynomials question
its in the 2001 HSC exam
Question 7 (b) all parts
Consider the equation x³ - 3x - 1 = 0, which we denote by (*).
(i) Let x = p/q where p and q are integers having no common divisors other than +1 and -1. Suppose that x is a root of the equation ax³ - 3x + b = 0, where a and b are integers.
Explain why p divides b and why q divides a. Deduce that (*) does not have a rational root. [4 marks]

(ii) Suppose that r, s and d are rational numbers and that sqrt(d) is irrational. Assume that [r + s x sqrt(d)] is a root of (*).
Show that 3r²s + s³d - 3s = 0 and show that [r - s x sqrt(d)] must also be a root of (*).
Deduce from this result and part (i), that no root of (*) can be expressed in the form r + s sqrt(d) with r, s and d rational. [4 marks]

(iii) Show that one root of (*) is 2cos(pi/9). (You may assume the identity cos3@ = 4cos³@ - 3cos@.) where @ is theta. [1 mark]

i can sorta do part (i) but i want to make sure i have the correct working out. thanx

 

[KFunk]

Consider the equation x³ - 3x - 1 = 0, which we denote by (*).
(i) Let x = p/q where p and q are integers having no common divisors other than +1 and -1. Suppose that x is a root of the equation ax³ - 3x + b = 0, where a and b are integers.
Explain why p divides b and why q divides a. Deduce that (*) does not have a rational root. [4 marks]

i can sorta do part (i) but i want to make sure i have the correct working out. thanx

Here's the way it makes sense to me. First off show that p divides b and q divides a so sub x = p/q into the equation:

ap³/q³ - 3p/q + b = 0
ap³ - 3pq² + bq³ = 0

bq³ = p(3q² - ap²) = pR₁ (where R₁ is an integer)

hence bq³/p = R₁ , but p does not divide q (as defined) so p divides b. Similarly:

ap³ = q(3pq - bq²) = qR₂ (where R₂ is an integer)

hence qp³/q = R₂, but q does not divide p, so q divides a.


Then assume that (*) has a rational root (p/q). In eq'n (*) a=1 and b=-1. Since q divides a, q = ±1 and since p divides b, p = ± 1. Hence if (*) has a rational root then p/q = ± 1:

Subbing 1 into (*) = -3
Subbing -1 into (*) = 1

We arrive at a contradiction since neither 'possible' rational root is a zero of (*), ∴ the roots of (*) are irrational.

 

[justchillin]

For the second part of the question, sub r + sqrt (d) or whatever it is into the equation. You get a rational part + an irrational part = 0 with correct factorising. Now, just like equating imaginary parts in complex numbers, equate the irrational part and u get the expression for that = 0. Also note the other expression as its needed next. To prove r -sqrt(d) is a root, sub in and play around with the expression, and u get sqrt(d) outside the expression that I told u to take note of. Now since this equals zero, by equating the rational parts of the other root, the whole thing = 0 and hence r- sqrt(d) is also a root. For the cosine part, let x=cos(theta) then let cos 3(theta)=1 and solve...