Question

Find an equation of a polynomial (in polynomial form) with real coefficients that has a degree of 4, y-

intercept of (0, 130), and has zeros of 1 - 2i, 3 – 2i.​

Answer 1

2x^4 - 16x^3 + 60x^2 - 112x + 130 = 0.

Explanation:

It has a total of 4 roots as it is of degree 4.

In addition to the 2 given complex roots we have their complements.

So the 4 roots are 1 - 2i, 1 + 2i, 3-2i and 3 + 2i.

So in factor form we have:

(x - (1 - 2i))(x - (1 + 2i))(x - (3 - 2i))(x - (3 + 2i)) = 0

(x - 1 + 2i))(x - 1 - 2i))(x - 3 + 2i))(x - 3 - 2i)) = 0

The first 2 parentheses simplify to:

(x^2 - x - 2ix - x + 1 + 2i + 2ix - 2i - 4i^2)

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

= x^2 - 2x + 5

Second 2 parentheses:

(x - 3 + 2i))(x - 3 - 2i)

= x^2 - 3x - 2ix - 3x + 9 + 6i + 2Ix - 6i - 4i^2

= x^2 - 6x + 9 + 4

= x^2 - 6x + 13

So we multiply:

(x^2 - 2x + 5)(x^2 - 6x + 13)

= x^4 - 6x^3 + 13x^2 - 2x^3 + 12x^2 - 26x + 5x^2 - 30x + 65

= x^4 - 8x^3 + 30x^2 - 56x + 65.

Now the polynomial passes through the point ( 0 , 130) so

a (4 *0 - 8*0 + 30*0^2 - 56*0 + 65) = 130

65a = 130

a =2.

So the required equation is

2(x^4 - 8x^3 + 30x^2 - 56x + 65) = 0

= 2x^4 - 16x^3 + 60x^2 - 112x + 130 = 0.