How do you find the zeros of #5x^8+4x^7+20x^5+42x^4+20x^3+4x+5# algebraically?

3 Answers
Konstantinos Michailidis
Jun 4, 2016

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Explanation:

We can factor this as a product of fourth degree polynomials

#5x^8+4x^7+20x^5+42x^4+20x^3+4x+5=(x^4+4 x+5) (5 x^4+4 x^3+1)#

The first #x^4+4x+5# can be factored as

#(-x+sqrt(1-i)-i) (-x+sqrt(1+i)+i) (x+sqrt(1-i)+i) (x+sqrt(1+i)-i) #

The second #5x^4+4x^3+1# can be factored as

#1/125 (-5 x+sqrt(7-i)-(1+2 i)) (-5 x+sqrt(7+i)-(1-2 i)) (5 x+sqrt(7-i)+(1+2 i)) (5 x+sqrt(7+i)+(1-2 i)) #

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

fourth degree polynomial solutions

George C.
Jun 11, 2016

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 #c != 0#, hence either #(a, c) = (1, -1)# or #(a, c) = (0, 4)#

If #(a, c) = (1, -1)# then we find #b^2=14#, which would make #b# not an integer.

If #(a, c) = (0, 4)# then we find #b=0#.

The values #a=b=0#, #c=4# satisfy all of the equations, so we have found our factorisation:

#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.

George C.
Jun 11, 2016

Part Two - Solving the first quartic

Find zeros of #x^4+4x+5# by an algebraic method...

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 #t=A^2#

We want to solve:

#t^3-20t-16 = 0#

By the rational root theorem, any rational roots are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-16# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1#, #+-2#, #+-4#, #+-8#, #+-16#

We find #t=-4# is a zero, so #(t+4)# is a factor:

#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 #t=2+2sqrt(2)# to get a factorisation into quadratics with real coefficients, but the sums may be a little easier by choosing #A^2=t=-4#.

Let #A=2i#

Then:

#{ (B + C = -4), (B - C = 4/(2i) = -2i) :}#

Adding we get:

#2B = -4-2i#

Hence #B=-2-i# and #C=-2+i#

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 #Arg(z) in (-pi, pi]#):

#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 #a+bi# form are:

#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#

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