cssa 2004 polynomials (need help) (1 Viewer)

[blackfriday]

question five. does anyone have answers for the paper? if not...

(a) the equation x^4 - x^3 +2x^2 -2x +1 =0 had roots a,b,c,d.

(i) show that none of a, b, c, d is an integer

(ii) find the monic equation of degree four with roots a-1, b-1, c-1, d-1, and hence find the value of (a+b+c)(b+c+d)(a+c+d)(a+b+d).

(b) (i) express the roots of the equation z^5+32=0 in mod/arg form.

(ii) hence show that z^4 -2z^3 +4^2 -8z +16 = {z^2 -(4cos(pi/5)z +4}{z^2 -(4cos(3pi/5)z +4}

(iii) hence find the exact values of cos(pi/5) and cos(3pi/5) in simplest surd form.

cheers guys. the guys i asked in my class are pretty stumped and we got a polynomials assesment on tuesday!

 

[Ogden_Nash]

(a)i) Sum and product of roots = 1. Therefore none of the roots can be integral. (Not 100% sure that reason is adequate)

ii) Replace x with x+1 in the original polynomial to yield the new polynomial. Now, notice that a+b+c = 1-d (sum of roots).
Therefore (a+b+c)(b+c+d)(a+c+d)(a+b+d) = (1-d)(1-a)(1-b)(1-c)
=-(d-1).-(a-1).-(b-1).-(c-1)
=(d-1)(a-1)(b-1)(c-1)
= product of roots of new equation

b) i) Find the five 5th complex roots of -32

ii) Equate the factored forms of the polynomial over the integers and over the reals

iii) Equate coefficients and form a quadratic with roots cos(pi/5) and cos(3pi/5) [You know the sum and product of the roots from equating coefficients]. Solve the quadratic using the quadratic formula and use the fact that cos(pi/5) > cos(3pi/5) to sort them out.

 

[blackfriday]

um thanx, but b part 2 is still a big confusing mess to me.

 

[Ogden_Nash]

z^5 - 32 = (z-2)(z^4 -2z^3 +4^2 -8z +16)

and

z^5 - 32 = (z-2){z^2 -(4cos(pi/5)z +4}{z^2 -(4cos(3pi/5)z +4}

 

[blackfriday]

oops i mean i dont get b part 3. i got part 1 and part 2 alright.