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write a polynomial function f(x), such that f(x) is a third-degree function with zeros at 1, 6, and -9

User Stivlo
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4 votes

Answer:

f(x) = x³ - 2x² - 57x + 54

Explanation:

given the zeros of f(x) are x = a , x = b, x = c , then the corresponding factors are (x - a) , (x - b) , (x - c)

then f(x) is the product of the factors , that is

f(x) = k(x - a)(x - b)(x - c) ← where k is a multiplier

here the zeros are x = 1, x = 6 and x = - 9 , then the factors are

(x - 1) , (x - 6) , (x - (- 9) ) , that is (x - 1) , (x - 6) , (x + 9) , then

f(x) = k(x - 1)(x - 6)(x + 9)

to find k we require some other point that lies on f(x)

in this case let k = 1 , so

f(x) = (x - 1)(x - 6)(x + 9) ← expand the first 2 factors using FOIL

= (x² - 7x + 6)(x + 9) ← distribute

= x³ + 9x² - 7x² - 63x + 6x + 54 ← collect like terms

f(x) = x³ + 2x² - 57x + 54

User VanDir
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