Answer:
2x^4 - 16x^3 + 60x^2 - 112x + 130 = 0.
Step-by-step 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.
