Find the Roots (Zeros) h(x)=4x^4-5x^3+2x^2-x+5

Problem

h(x)=4*x4−5*x3+2*x2−x+5

Solution

1. Identify the degree of the polynomial. Since the highest power of x is 4 the Fundamental Theorem of Algebra states there are exactly 4 roots (including real and complex roots).

2. Apply the Rational Root Theorem to find potential rational roots. The possible rational roots are factors of the constant term 5 divided by factors of the leading coefficient 4

Possible roots=±1,±5,±1/2,±5/2,±1/4,±5/4

3. Test the potential rational roots using synthetic division or substitution.

h(1)=4*(1)4−5*(1)3+2*(1)2−(1)+5=5

h*(−1)=4*(−1)4−5*(−1)3+2*(−1)2−(−1)+5=17

Testing other rational candidates like 5/4 or −5/4 also yields non-zero results, suggesting no rational roots exist.

4. Analyze the function using Descartes' Rule of Signs. There are 4 sign changes in h(x) indicating 4 2 or 0 positive real roots. For h*(−x)=4*x4+5*x3+2*x2+x+5 there are 0 sign changes, indicating 0 negative real roots.

5. Evaluate the discriminant or use numerical methods. Since there are no rational roots and the function h(x) remains positive for all real x (the minimum value of the quartic is approximately 4.58, all roots must be complex.

6. Solve for the complex roots using the quartic formula or numerical approximation. The roots occur in two conjugate pairs.

x≈0.701±0.781*i

x≈−0.076±0.915*i

Final Answer

x≈0.701±0.781*i,−0.076±0.915*i

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