Question

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

Answer 1

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`