• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Polynomials Q: help needed pls! (1 Viewer)

jjnfg

New Member
Joined
Nov 22, 2003
Messages
4
I have a problem with these questions; if anyone knows how to do it I would be very grateful!!

1) If the sum of two roots of x (to the power 4) + 2x (to the power 3) - 8x (squared) - 18x - 9=0 is 0, find the roots of the equation.

2) Solve 2cos (cubed)x + cos (squared) x - cosx =0 for x is more than or equal to o degrees and less or equal to 360 degrees.

Thanks heaps.
 

Harimau

Member
Joined
Mar 24, 2003
Messages
430
Location
You are not me, therefore you are irrelevant
Gender
Male
HSC
2003
1) Can't be bothered to such long things. Just let the roots be A, -A, B and C. From then on use the sum of roots, the product of roots, the sum of roots three at a time, the sum of roots two at a time, and use simultaneous equations.

2) Factorise the Cosx:

cosx (2cos^2x + cosx - 1) =0
cosx (2cosx -1)(cosx+1)=0

Solve normally from there for cosx= 0, cosx=1/2, cosx= -1 with attention to the domain. Make sure your end result is between 0 and 360 degrees, and they will all be exact value as well.
 

Toodulu

werd!
Joined
Apr 15, 2003
Messages
1,335
Location
Sydney
Gender
Female
HSC
2003
for the first one, i tested the divisors of the constant term, -1 and -3 were roots
i did P(x) = (x+3) (x+1) (x -A) (x-B) where A and B are the 2 other roots
so P(x) = (x^2+4x+3) (x^2-2x-3) by inspection
and then i factorised it
ie. P(x) = (x+3)(x+1)(x-3)(x+1)
roots are 3, -3, -1, -1
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top