How do you find the zeros of #5x^8+4x^7+20x^5+42x^4+20x^3+4x+5# algebraically?
3 Answers
bolded text
Explanation:
We can factor this as a product of fourth degree polynomials
The first
The second
Now we can easily describe the zeros of the original equation.
Footnote
For general solutions of fourh degree equations the following article may be useful
Part one: Factoring into quartics
#5x^8+4x^7+20x^5+42x^4+20x^3+4x+5#
#= (x^4+4x+5)(5x^4+4x^3+1)#
Explanation:
#f(x) = 5x^8+4x^7+20x^5+42x^4+20x^3+4x+5#
Part One - Factoring into quartics
Note that since the coefficients are symmetric, this can be factored as a product of quartics with coefficients in mirror image to one another.
So there is a factorisation of the form:
#5x^8+4x^7+20x^5+42x^4+20x^3+4x+5#
#=(x^4+ax^3+bx^2+cx+5)(5x^4+cx^3+bx^2+ax+1)#
#=5x^8+(5a+c)x^7+(ac+6b)x^6+(a+ab+bc+5c)x^5+(a^2+b^2+c^2+26)x^4+(a+ab+bc+5c)x^3+(ac+6b)x^2+(5a+c)x+5#
Equating coefficients, we find:
#{ (5a+c = 4), (5b+ac+b = 0), (a+ab+bc+5c = 20), (a^2+b^2+c^2+26 = 42) :}#
From the last equation, we find:
#a^2+b^2+c^2 = 16#
and hence:
#{ (abs(a) <= 4), (abs(b) <= 4), (abs(c) <= 4) :}#
If there are integer solutions, then the first equation implies
If
If
The values
#5x^8+4x^7+20x^5+42x^4+20x^3+4x+5#
#=(x^4+4x+5)(5x^4+4x^3+1)#
The zeros of the second quartic will be the reciprocals of the first one.
Part Two - Solving the first quartic
Find zeros of
Explanation:
Part Two - Solving the first quartic
#g(x) = x^4+4x+5#
#= (x^2-Ax+B)(x^2+Ax+C)#
#= x^4 + (B+C-A^2)x^2 + A(B-C)x + BC#
Equating coefficients and rearranging a little:
#{ (B + C = A^2), (B - C = 4/A), (BC = 5) :}#
So:
#(A^2)^2 = (B+C)^2 = (B-C)^2 + 4BC = 16/A^2 + 20#
Hence:
#(A^2)^3-20(A^2)-16 = 0#
Let
We want to solve:
#t^3-20t-16 = 0#
By the rational root theorem, any rational roots are expressible in the form
That means that the only possible rational roots are:
#+-1# ,#+-2# ,#+-4# ,#+-8# ,#+-16#
We find
#t^3-20t-16 = (t+4)(t^2-4t-4)#
#=(t+4)((t-2)^2-(2sqrt(2))^2)#
#=(t+4)(t-2-2sqrt(2))(t-2+2sqrt(2))#
We could use the positive root
Let
Then:
#{ (B + C = -4), (B - C = 4/(2i) = -2i) :}#
Adding we get:
#2B = -4-2i#
Hence
So we have found:
#x^4+4x+5 = (x^2-2ix-2-i)(x^2+2ix-2+i)#
So using the quadratic formula on the first of these quadratic factors we find:
#x = (2i+-sqrt((2i)^2+4(2-i)))/2 = (2i+-sqrt(4-4i))/2 = i+-sqrt(1-i)#
The second quadratic is the Complex conjugate, so we can deduce zeros:
#x = -i+-sqrt(1+i)#
Next use the formula I derived in https://socratic.org/s/avemtNYy , which simplified for square roots in Q1 becomes:
#sqrt(a+bi) = (sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i#
So:
#sqrt(1+i) = (sqrt((sqrt(1^2+1^2)+1)/2)) + (sqrt((sqrt(1^2+1^2)-1)/2))i#
# = (sqrt((sqrt(2)+1)/2))+(sqrt((sqrt(2)-1)/2))i#
Hence (using the normal definition of
#sqrt(1-i) = (sqrt((sqrt(2)+1)/2))-(sqrt((sqrt(2)-1)/2))i#
So:
#i +-sqrt(1-i) = i +-((sqrt((sqrt(2)+1)/2))-(sqrt((sqrt(2)-1)/2))i)#
#-i+-sqrt(1+i) = -i +-((sqrt((sqrt(2)+1)/2))+(sqrt((sqrt(2)-1)/2))i)#
So the four zeros in
#x_1 = (sqrt((sqrt(2)+1)/2))-(1+sqrt((sqrt(2)-1)/2))i#
#x_2 = -(sqrt((sqrt(2)+1)/2))-(1-sqrt((sqrt(2)-1)/2))i#
#x_3 = (sqrt((sqrt(2)+1)/2))-(1-sqrt((sqrt(2)-1)/2))i#
#x_4 = -(sqrt((sqrt(2)+1)/2))-(1+sqrt((sqrt(2)-1)/2))i#