How do you find the zeros of # 4x^4 + 3x^3 + 2x^2 - 3x + 4#?

1 Answer
Feb 17, 2016

Answer:

See explanation...

Explanation:

This is an interesting question due to the form of the quartic.

First note the symmetry/anti-symmetry of the coefficients. This results in the property that if #x=r# is a zero then #x=-1/r# is also a zero:

Let #f(x) = 4x^4+3x^3+2x^2-3x+4#

Then:

#r^4 f(-1/r) = r^4 (4r^(-4)-3r^(-3)+2r^(-2)+3r^(-1)+4)#

#= 4-3r+2r^2+3r^3+4r^4 = f(r)#

Taking such a pair of zeros, we find:

#(x-r)(x+1/r) = x^2+(1/r-r)x-1#

So #f(x)# will have a quadratic factorization of the form:

#4x^4+3x^3+2x^2-3x+4#

#=4(x^2+(1/r_1-r_1)x-1)(x^2+(1/r_2-r_2)x-1)#

#=(2x^2+2(1/r_1-r_1)x-2)(2x^2+2(1/r_2-r_2)x-2)#

Therefore consider:

#4x^4+3x^3+2x^2-3x+4#

#= (2x^2+ax-2)(2x^2+bx-2)#

#= 4x^4+2(a+b)x^3+(ab-8)x^2-2(a+b)x+4#

Equating coefficients we find:

#2(a+b) = 3#

#ab = 10#

Hence (to cut a long story a little shorter) we can find:

#a=3/4+sqrt(151)/4i#

#b=3/4-sqrt(151)/4i#

So the zeros of #f(x)# are the roots of the two quadratic equations:

#2x^2+(3/4+sqrt(151)/4i)x-2 = 0#

#2x^2+(3/4-sqrt(151)/4i)x-2 = 0#

Using the quadratic formula to find the roots, we obtain formulae including the square root of a Complex number.

#x = 1/16(-3-sqrt(151)i+-sqrt(114+6sqrt(151)i))#

#x = 1/16(-3+sqrt(151)i+-sqrt(114-6sqrt(151)i))#

To convert these to #a+bi# form use the method described in http://socratic.org/questions/how-do-you-find-the-square-root-of-an-imaginary-number-of-the-form-a-bi

The square roots of #c+di# are:

#+-((sqrt((sqrt(c^2+d^2)+c)/2)) + (d/abs(d) sqrt((sqrt(c^2+d^2)-c)/2))i)#

Hence (using principal square root based on #Arg(z) in (-pi, pi]#):

#sqrt(114+6sqrt(151)i) = sqrt(48sqrt(2)+57) + (sqrt(48sqrt(2)-57))i#

#sqrt(114-6sqrt(151)i) = sqrt(48sqrt(2)+57) - (sqrt(48sqrt(2)-57))i#