<span>x^3 - 6x^2 + 13x - 10
We have been given the roots 2+i and 2, so two of the factors in the desired polynomial is
(x - 2) = 0
(x - 2 -i ) = 0
Let's create a third root by negating the sign of the imaginary component to create the complex conjugate of the known complex root.
(x - 2 + i) = 0
Now multiple (x - 2 - i) and (x - 2 + i)
(x - 2 - i)
(x - 2 + i)
x^2 - 2x - xi
-2x +4 +2i
xi -2i - (-1)
x^2 -4x +4 + 1
x^2 -4x + 5
And we've cancelled the imaginary term. Let's multiply this quadratic equation by the remaining factor.
x^2 -4x + 5
x - 2
x^3 - 4x^2 + 5x
-2x^2 +8x -10
= x^3 -6x^2 + 13x - 10
So the desired polynomial is x^3 - 6x^2 + 13x - 10</span>
I'm pretty sure it's figure A
Triangle jkl is right of triangle what is the length of JK
Answer:
C.
Step-by-step explanation:
c is the only one that even makes sense
