Question 409907: dividing polynomies
Hi!
I got my HW and i don't know how to do it and y need help
my problems are..
81x3-192=0
125x3+216=0
x4-64=0
-2x4+46x2=-100
27=-x4-12x2
x5-5x3+4x=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 81x3-192=0
125x3+216=0
x4-64=0
-2x4+46x2=-100
27=-x4-12x2
x5-5x3+4x=0
..
81x3-192=0
81x^3=192
x^3=192/81
take the cube root of both sides,
x=cube root(192/81)
=cube root(64*3/27*3)
=4cube root(3)/3cube root(3)
=4/3
..
125x3+216=0
125x^3+216=0
125x^3=-216
take the cube root of both sides
5x=-6
x=-6/5
..
x4-64=0
x^4=64
take the fourth root of both sides
x=64^1/4
=(2^4*4)^1/4
=2*4^1/4
..
-2x4+46x2=-100
-2x^4+46x^2=-100
2x^4-46x^2=100
2x^4-46x^2-100=0
x^4-23x^2-50=0
solve for x by factoring
(x^2-25)(x^2+2)=0
x^2-25=0
x=+-sqrt(25)
x=5
x=-5
x^2+2=0
x=+-sqrt(2)
x=sqrt(2)
x=-sqrt(2)
The four roots are:5,-5,sqrt(2),-sqrt(2)
..
27=-x4-12x2
x^4+12x^2+27=0
factor
(x^2+9)(x^2+3)
x^2+9=0
x=+-sqrt(9)
x=+-3
x^2+3=0
x=+-sqrt(3)
The four roots are:3,-3,sqrt(3),-sqrt(3)
..
x5-5x3+4x=0
x^5-5x^3+4x=0
factor out x
x(x^4-5x^2+4)=0
x(x^2-4)(x^2-1)=0
x=0
x^2-4=0
x=+-sqrt(4)
x=+-2
x^2-1=0
x=sqrt(1)
x=+-1
The five roots are:0,-2,2,-1,1
Hope this helps! Be sure to continue doing your homework.
|
|
|
