f ( x ) = x 3 − 5 x 2 + 4 x + 10 f ( x ) = x 3 + x 2 − 6 x 2 − 6 x + 10 x + 10 f ( x ) = x 2 ( x + 1 ) − 6 x ( x + 1 ) + 10 ( x + 1 ) f ( x ) = ( x + 1 ) ( x 2 − 6 x + 10 ) x + 1 = 0 x = − 1 x 2 − 6 x + 10 = 0 x 2 − 6 x + 9 + 1 = 0 ( x − 3 ) 2 = − 1 x − 3 = i ∨ x − 3 = − i x = 3 + i ∨ x = 3 − i x = { − 1 , 3 − i , 3 + i }
The zeros of the function f ( x ) = x 3 − 5 x 2 + 4 x + 10 are − 1 , 3 + i , and 3 − i . We found one real root through substitution and then factored the polynomial to find the complex roots. Using the quadratic formula, we calculated the complex roots from the remaining quadratic equation.
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