Determine the number and type of solutions for the following polynomial function. Make sure to include an explanation in your work to justify how you determined your answer.
P(x)= -2x^4 + 2x^2 + 4
A. 4 real solutions, 0 imaginary solutions
B. 2 real solutions, 0 imaginary solutions
C. 2 real solutions, 2 imaginary solutions
D. 1 real solution, 3 imaginary solutions
To determine the number and types of solutions for the polynomial function P(x) = -2x^4 + 2x^2 + 4, we can look at the degree of the polynomial and the sign of the leading coefficient, as well as try to find its roots if possible.
Firstly, we know that a fourth-degree polynomial can have up to four real solutions (roots), and by the Fundamental Theorem of Algebra, it will have exactly four solutions when counting multiplicity and including both real and non-real (complex) roots.
Let's try to simplify and factor the polynomial if possible to find the roots:
P(x) = -2x^4 + 2x^2 + 4
We can factor out a 2 from all terms:
2(-x^4 + x^2 + 2)
It's not readily factorable, but we can try a substitution: Let u = x^2. The equation then becomes a quadratic in terms of u:
-2u^2 + 2u + 4
Divide everything by -2 for simplification:
u^2 - u - 2
Now, we can factor this quadratic polynomial:
(u - 2)(u + 1)
So the solutions for u are u = 2 and u = -1.
Let's go back to x:
x^2 = 2 --> x = √2 or x = -√2
x^2 = -1 --> x is imaginary since there are no real numbers that square to give -1.
Since we have two possible values for x from u = 2 (which are real) and none from u = -1 (which are imaginary), we have:
- Two real roots from x = √2 and x = -√2.
- Two imaginary roots from x = i (the square root of -1) and x = -i.
Hence, the correct answer is:
C. 2 real solutions, 2 imaginary solutions.