I dont know if the work is right but this is kinda how it wa

I dont know if the work is right, but this is kinda how it was taught to us. Pleaseee Helppp!

3. Find all complex-zeros of the o polynomial fnction f(a)14-+2-16 d all complex heros of the

Solution

f(x) = x^5 - 6x^4 +14x^3 -20x^2 +24x -16

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction pq, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient (1) is 1 .The factors of the constant term (-16) are 1 2 4 4 8 16 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±4/1, ±4/1, ±8/1, ±16/1

Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(2)=0.

(x^5 - 6x^4 +14x^3 -20x^2 +24x -16) /(x-2) = x^44x^3+6x^28x+8

Again repeat the above process:

The factor of the leading coefficient (1) is 1 .The factors of the constant term (8) are 1 2 4 8 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±4/1, ±8/1

Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(2)=0.

(x^44x^3+6x^28x+8)/(x -2) = x^32x^2+2x4

Repeat the same process:

The factor of the leading coefficient (1) is 1 .The factors of the constant term (-4) are 1 2 2 4 . Then the Rational Roots Tests yields the following possible solutions:

±11, ±21, ±21, ±41

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(2)=0.

(x^32x^2+2x4)/(x-2) = x^2 +2

Roots of x^2 +2 are x = i*sqrt2 , -i*sqrt2

So, complex roots are x =  i*sqrt2 , -i*sqrt2