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## 1 Answer

Find all zeros (real or complex) of f(x) = x^5-4x^4+4x^3+10x^2-13x-14:

x^5-4x^4+4x^3+10x^2-13x-14 = 0

By inspection x = -1 is a root, hence (x+1) is a factor. Taking out this factor,

(x+1)(x^4 - 5x^3 + 9x^2 + x - 14) = 0

By inspection x = -1 is again a root, hence (x+1) is again a factor. Taking out this factor,

(x+1)(x+1)(x^3 - 6x^2 + 15x - 14) = 0

By inspection x = 2 is root, hence (x-2) is a factor. Taking out this factor,

(x+1)^2(x-2)(x^2 - 4x + 7) = 0

Using the quadratic formula to solve the quadratic equation,

x = (4 +/ sqrt(16 - 28))/(2) = 2 +/- sqrt(4 - 7)

x = 2 +/- i.sqrt(3)

The five roots then are: x = -1 (twice), x = 2, x = 2 +/- i.sqrt(3)

by Level 11 User (80.8k points)

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