How do you find all the zeros of $f (x) = x^{3} - x^{2} + 25 x - 25$ $f(x) = x^3-x^2+25x-25$ ?

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Answer 1 · 2016-04-25T06:24:57

$x = 1, \pm 5 i$ $x = 1, +-5 i$ .

Sum of the coefficients = 0. So, 1 is a root.

One factor of the cubic is $(x - 1) .$ $(x - 1).$ .

The other quadratic factor is readily seen as $(x^{2} + 5^{2})$ $(x^2+5^2)$ .

$(x^{2} + 5^{2}) = 0$ $(x^2+5^2) = 0$ gives the other two roots as $\pm 5 i$ $+- 5 i$ .

How do you find all the zeros of f(x) = x^3-x^2+25x-25f(x)=x3−x2+25x−25 f(x) = x^3-x^2+25x-25?