Find all other zeros of P(x)= x^3 - 6x^2 +16x - 16 given that 2+2i is a zero.?

If 2+2i is a zero, then 2-2i is also a zero (as complex zeros come in conjugate pairs.

Multiply out

(x-(2+2i))(x-(2-2i))

= x^2 - x(2-2i) - x(2+2i) + (2+2i)(2-2i)

= x^2 - 2x + 2ix - 2x - 2ix + 4 - 4i + 4i - 4i^2

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

= x^2 - 4x + 8

Now use long division.

(x^3 -6x^ +16x -16) / (x^2 -4x +8) = x - 2

Set it equal to zero.

x-2 = 0 when x = 2

That means 2 is the other zero.

If 2 + 2i is a zero, so is 2 - 2i, so both (x - 2 - 2i) and (x - 2 + 2i) are factors, and (x - 2 - 2i)(x - 2 + 2i) = x² - 4x + 8 divides x^3 - 6x² + 16x - 16, leaving x - 2 as a factor and 2 as a zero,

so the zeros are 2, 2-2i, and 2+2i

First of all, conjugates appear in pairs, so if 2+2i is a zero then so is 2-2i.

The function can be written as:

f(x)=(x-(2+2i))(x-(2-2i))(x-a)

a is the the root to be found.

The -16 term in the equation is made up of

-(2+2i)*-(2-2i)*-a = -16

-8a = -16

Sooo a = 2.

The missing zero is x = 2.

The complex non real roots of a polynomila with real coeeficients are always cojugate, so that, 2 - 2i is another root of P. According to the Girad relations, the sum of the roots of P is the symmetric of the ratio between the coeeficient of x^2 (the term with degree 1 unit less the degree of the leading term) and the coefficient of x^3, the leading term. So, the sum of the 3 roots (P has degree 3) is r + 2 + i + 2 -i = -(-6)/1 = 6, so that r = 2.

In fact, we readily see that P(r) = 0.

x1=2+2i =>

x2=2-2i is a solution too

so x1+x2=4

x1*x2=(2+2i)(2-2i)=4+4=8

=> we divide x^3-6^x^2+16^x-16 by x^2-4x+8 => x-2

x+2=0 => x3=-2

or using Matlab:

v=[1 -6 16 -16 ]

( the coefficients of the equation in the proper order, from the highest power to the lowest one )

roots(v)

that leads us to:

2+2i, 2-2i,2

or division using matlab:

a=[1 -6 16 -16 ]

b=[1 -4 8]

deconv(a,b)

=> 1 -2 ( aka x-2 )

Complex zeros always occur in conjugate pairs. Thus another zero is immediate, namely, 2 - 2i

Thus(x - 2 +2i)(x +2 +2i) is a factor and the rest is easy!