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Find a polynomial f (X) of degree 3 with real coefficients and the following zero 4, 2 - i

User Nahtnam
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1 Answer

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Answer:

f(x) = x³ - 8x² + 21x - 20

Explanation:

∵ x = a + bi is a root of f(x)

∴ x = a - bi is the second root of f(x)

∴ f(x) = x² - (sum of the roots)x + (the product of the roots)

∵ 2 - i is a root of f(x)

2 + i is the other root of f(x) ⇒ (conjugate to each other)

∵ The sum of the roots = 2 - i + 2 + i = 4

∵ The product of them = (2 - i)(2 + i) = 4 + 2i - 2i - i² = 4 + 1 = 5⇒i² = -1

∴ f(x) = x² - 4x + 5

∵ 4 is a root of f(x)

x - 4 is a factor of f(x)

∴ f(x) = (x² - 4x + 5)(x - 4) = x³ - 4x² - 4x² + 16x + 5x - 20

∴ f(x) = x³ - 8x² + 21x - 20

User Sidra
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