factoring and zeros?

Question

can some body help me?
the question is
i is a zero of y= x^4 -4x³ + 2x² -4x + 1
list all the zeros?

please explain . any help is appreciated. thnx~! ^-^

Answer 1

If i is a factor, then -i is also going to be a factor.

(x - i)(x + i) = (x² + ix - ix - i²) = x² + 1

So use synthetic division to divide this by x² + 1:
................... .................... .. x² - 4x + 1
x² + 0x + 1 ) x^4 - 4x^3 + 2x² - 4x + 1
................... x^4 + 0x^3 + x²
........................... -4x^3 + x² - 4x
........................... -4x^3 + 0 - 4x
................... .................... x² + 0x + 1
................... .................... x² + 0x + 1

This gives you:
(x² + 1)(x² - 4x + 1)

Using the quadratic formula on the second half:
a = 1, b = -4, c = 1
x = [ -(-4) ± sqrt( (-4)² - 4(1)(1) ) ] / 2
x = [ 4 ± sqrt( 16 - 4 ) ] / 2
x = [4 ± sqrt(12) ] / 2
x = 2 ± sqrt(3)

So your roots are:
i, -i, 2 + √3, 2 - √3

Answer 2

If i is a zero, then so is -i.

So (x^2 + 1) is one of the factors of the expression.
[x^2 + 1 = 0 ---> x = ±√-1]

So with long division, divide the expression by x^2 + 1

............ ...............x^2 - 4x + 1
.............___________________
x^2 + 1 | x^4 - 4x^3 + 2x^2 - 4x + 1
............ ..x^4 +... ....+x^2
............ ..-4x^3 +x^2 - 4x
............ ..-4x^3 + ..... - 4x
............ .. x^2 + 1
............ .. x^2 + 1
............ .. 0


So
x^4 -4x³ + 2x² -4x + 1
= (x^2 - 4x + 1)(x^2 + 1)

x^2 - 4x + 1 = 0
Use the quadratic formula:
x = [4 ± sqrt(16 - 4)] / 2
x = [4 ± sqrt(12)] / 2
x = [4 ± 2sqrt(3)] / 2
x = 2 ± sqrt(3)

So the four zeros, both real and complex, are:
x = -i, i, 2-√3, 2+√3