Find roots of polynomial $p (x) = - x^{5} + 4 x^{4} + x^{3} - 15 x^{2} + 12 x + 4$

Answer

The roots of polynomial $p (x)$ are:
$x_{1} x_{2} x_{3} x_{4} x_{5} = - 0.2533 = - 1.9295 = 3.2888 = 1.447 + 0.6287 i = 1.447 - 0.6287 i$

Explanation

Polynomial $- x^{5} + 4 x^{4} + x^{3} - 15 x^{2} + 12 x + 4$ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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Find roots of polynomial p(x)=−x5+4x4+x3−15x2+12x+4

The roots of polynomial p(x) are: x1​x2​x3​x4​x5​​=−0.2533=−1.9295=3.2888=1.447+0.6287i=1.447−0.6287i​

Find roots of polynomial $p (x) = - x^{5} + 4 x^{4} + x^{3} - 15 x^{2} + 12 x + 4$

The roots of polynomial $p (x)$ are:
$x_{1} x_{2} x_{3} x_{4} x_{5} = - 0.2533 = - 1.9295 = 3.2888 = 1.447 + 0.6287 i = 1.447 - 0.6287 i$