x^3+4x^2+5x-1=0

Question

i need to know the complex root, posible real numers, and possible rational roots

Answer 1

This cubic has one real root and two complex roots.

There are several ways to find the real root. I find Newton's iterative method the best.

Let f(x)=x³+4x²+5x-1, f'(x)=3x²+8x+5;

x_n+1=x_n-f(x_n)/f'(x_n), where x₀=0.

x₁=⅕, x₂=7/40, x₃=0.1745595..., x₄=0.1745594103, x₅=x₄ (approx), so x=0.1745594103 (approx) is the real root.

To find the complex roots (conjugates) we need to reduce the cubic to a quadratic by dividing by the real root (synthetic division):

x | 1 4              5                    -1

     1 x       x²+4x     | x³+4x²+5x

     1 4+x  x²+4x+5  |        0

The coefficients of the quadratic are 1, 4.1745594103, 5.728708629.

These are respectively a, b and c for the quadratic formula:

x=(-4.1745594103±√(4.1745594103²-22.91483452))/2=-4.17456±2.34262i approx.