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

f(x) = x³ - 8x² + 21x - 20

Step-by-step 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

your answer is a

step-by-step explanation:

4*4*4*4=256

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: