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Using the given zero, find one other zero of f(x).

2 - 4i is a zero of f(x).= x4 - 4x3 + 21x2 - 4x + 20

1 + i
1
1 - i
2 + 4i

Respuesta :

zrh2sfo

Answer:

x = i or x = -i or x = 2 + 4 i or x = 2 - 4 i

Step-by-step explanation:

Solve for x:

x^4 - 4 x^3 + 21 x^2 - 4 x + 20 = 0

The left hand side factors into a product with two terms:

(x^2 + 1) (x^2 - 4 x + 20) = 0

Split into two equations:

x^2 + 1 = 0 or x^2 - 4 x + 20 = 0

Subtract 1 from both sides:

x^2 = -1 or x^2 - 4 x + 20 = 0

Take the square root of both sides:

x = i or x = -i or x^2 - 4 x + 20 = 0

Subtract 20 from both sides:

x = i or x = -i or x^2 - 4 x = -20

Add 4 to both sides:

x = i or x = -i or x^2 - 4 x + 4 = -16

Write the left hand side as a square:

x = i or x = -i or (x - 2)^2 = -16

Take the square root of both sides:

x = i or x = -i or x - 2 = 4 i or x - 2 = -4 i

Add 2 to both sides:

x = i or x = -i or x = 2 + 4 i or x - 2 = -4 i

Add 2 to both sides:

Answer:  x = i or x = -i or x = 2 + 4 i or x = 2 - 4 i

mberisso

Answer:

2 + 4i

Step-by-step explanation:

Given that the polynomial in question ( [tex]x^4-4x^3+21x^2-4x+20[/tex]) has all Real coefficients, if one of its zeros (roots) is the complex number "2 - 4i", for sure the "complex conjugate" of this complex root has to be a root as well. That is the only way that a complex root wouldn't show a non-real coefficient in the final polynomial.

So the complex root: "2 + 4i" has to be another zero.

This can be checked by evaluating this complex number in the original polynomial

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