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

Feb 17, 2016

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 $x = - \frac{1}{r}$ is also a zero:

Let $f \left(x\right) = 4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 3 x + 4$

Then:

${r}^{4} f \left(- \frac{1}{r}\right) = {r}^{4} \left(4 {r}^{- 4} - 3 {r}^{- 3} + 2 {r}^{- 2} + 3 {r}^{- 1} + 4\right)$

$= 4 - 3 r + 2 {r}^{2} + 3 {r}^{3} + 4 {r}^{4} = f \left(r\right)$

Taking such a pair of zeros, we find:

$\left(x - r\right) \left(x + \frac{1}{r}\right) = {x}^{2} + \left(\frac{1}{r} - r\right) x - 1$

So $f \left(x\right)$ will have a quadratic factorization of the form:

$4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 3 x + 4$

$= 4 \left({x}^{2} + \left(\frac{1}{r} _ 1 - {r}_{1}\right) x - 1\right) \left({x}^{2} + \left(\frac{1}{r} _ 2 - {r}_{2}\right) x - 1\right)$

$= \left(2 {x}^{2} + 2 \left(\frac{1}{r} _ 1 - {r}_{1}\right) x - 2\right) \left(2 {x}^{2} + 2 \left(\frac{1}{r} _ 2 - {r}_{2}\right) x - 2\right)$

Therefore consider:

$4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 3 x + 4$

$= \left(2 {x}^{2} + a x - 2\right) \left(2 {x}^{2} + b x - 2\right)$

$= 4 {x}^{4} + 2 \left(a + b\right) {x}^{3} + \left(a b - 8\right) {x}^{2} - 2 \left(a + b\right) x + 4$

Equating coefficients we find:

$2 \left(a + b\right) = 3$

$a b = 10$

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

$a = \frac{3}{4} + \frac{\sqrt{151}}{4} i$

$b = \frac{3}{4} - \frac{\sqrt{151}}{4} i$

So the zeros of $f \left(x\right)$ are the roots of the two quadratic equations:

$2 {x}^{2} + \left(\frac{3}{4} + \frac{\sqrt{151}}{4} i\right) x - 2 = 0$

$2 {x}^{2} + \left(\frac{3}{4} - \frac{\sqrt{151}}{4} i\right) x - 2 = 0$

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

$x = \frac{1}{16} \left(- 3 - \sqrt{151} i \pm \sqrt{114 + 6 \sqrt{151} i}\right)$

$x = \frac{1}{16} \left(- 3 + \sqrt{151} i \pm \sqrt{114 - 6 \sqrt{151} i}\right)$

To convert these to $a + b i$ 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 $c + \mathrm{di}$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{c}^{2} + {d}^{2}} + c}{2}}\right) + \left(\frac{d}{\left\mid d \right\mid} \sqrt{\frac{\sqrt{{c}^{2} + {d}^{2}} - c}{2}}\right) i\right)$

Hence (using principal square root based on $A r g \left(z\right) \in \left(- \pi , \pi\right]$):

$\sqrt{114 + 6 \sqrt{151} i} = \sqrt{48 \sqrt{2} + 57} + \left(\sqrt{48 \sqrt{2} - 57}\right) i$

$\sqrt{114 - 6 \sqrt{151} i} = \sqrt{48 \sqrt{2} + 57} - \left(\sqrt{48 \sqrt{2} - 57}\right) i$