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Published byNevaeh Leaton Modified over 10 years ago
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EXAMPLE 1 Find the number of solutions or zeros a. How many solutions does the equation x3 + 5x2 + 4x + 20 = 0 have? SOLUTION Because x3 + 5x2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)
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EXAMPLE 1 Find the number of solutions or zeros b. How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27 have? SOLUTION Because f (x) = x4 – 8x3 + 18x2 – 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)
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GUIDED PRACTICE for Example 1 1. How many solutions does the equation x4 + 5x2 – 36 = 0 have? ANSWER 4
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GUIDED PRACTICE for Example 1 2. How many zeros does the function f (x) = x3 + 7x2 + 8x – 16 have? ANSWER 3
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