Question

find the zeros of the biquadratic polynomial x^4-2x^3+4x^2+7x-5

Answer 1

${x}^{4} - 2 {x}^{3} + 4 {x}^{2} + 7x - 5 = 0$

The above quartic do not have rational solutions.

so

Using the Quartic Formula:

$\\ x_{1,2} = \frac{1 \pm \sqrt{ \frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } \pm i \sqrt{\sqrt{ \frac{1}{\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } }} \sqrt{ 5\sqrt{\frac{ - 5}{3} - \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } - \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } }+ (\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } )\sqrt{\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } \pm20} }{2}$

$\\ x_{3,4} = \frac{1 \pm \sqrt{ \frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } \mp i \sqrt[4]{ \frac{1}{\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } } \sqrt{ \frac{ - 5}{3} - \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } - \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } + (\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } )\sqrt{\frac{ - 5}{3} + \frac{1}{3} \sqrt[3]{ \frac{2855 + 3 \sqrt{905673} }{2} } + \frac{1}{3} \sqrt[3]{ \frac{2855 - 3 \sqrt{905673} }{2} } } \pm20} }{2}$

Slide to see the full length answer→→

I couldn't really make it look more simplified than this.

After approximation

x₁ ≈ − 1.3040

x₂ ≈ 0.56761

x₃ ≈ 1.3628 – 2.2098i

x₄ ≈ 1.3682 + 2.2098i

where, i = √(–1)