## anonymous 4 years ago Find the real and complex zeros of the following function. f(x)=x^3-6x^2+21x-26

Some possible roots for this function are$\pm 1, \pm 2, \pm 13, \pm 26$since those are the possible factors of the constant, 26, divided by the possible factors of the leading coefficient, 1. Using synthetic division, you can test those out to see which ( x- a) is a factor. You'll see that 2 works. So now the polynomial gets factored into:$f(x) = (x-2)(x^2 - 4x +13)$ Inspecting the quadratic will tell you that it is irreducible, you can have to solve it via the quadratic formula. Do so and you will see that your three solutions are:$x =2, x = 2 \pm 3i$