If you ever have a polynomial with a solution of bi, then it will also have a solution of -bi. Imaginary solutions always come in pairs.
So, yes, you could have a polynomial with solutions 2i and 5i, as long as -2i and -5i are also solutions.
(x-2i)(x+2i)(x-5i)(x+5i) would be the most basic polynomial you could form.
(x-2i)(x+2i)(x-5i)(x+5i) = (x^2+4)(x^2+25)
= x^4 + 29x^2 + 100
So the equation would be x^4 + 29x^2 + 100 = 0.
Now, if the question was "only the solutions of 2i and 5i and no others," then the answer is no, for the previously stated reason.
Where is the picture of the number line?
It is b because idk im just doing this so i can ask more questions
Answer:
h = -2
Step-by-step explanation:
1. Simplify the left hand side: 8h = -16
2. DIvide both sides by 8: h = -16 ÷ 8 = -2
Hope this helps!
The point (2, -1 , -8) is on the line and the expression (3, 9, 1) represents the family of vectors parallel to the line.
<h3>How to find a point on a line and a vector parallel to the line</h3>
First, we have to find the vector expression of the line function based on the following parametric equations:
t = (x - 2) / 3 (1)
x = 3 · t + 2
t = (y + 1) / 9 (2)
y = 9 · t - 1
t = z + 8 (3)
z = t - 8
(x, y, z) = (2, - 1, - 8) + t · (3, 9, 1) (4)
The point (2, -1 , -8) is on the line and the expression (3, 9, 1) represents the family of vectors parallel to the line.
To learn more on vectors: brainly.com/question/13322477
#SPJ1
