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. \[ P(x)=-2 x^{4}+2 x^{2}+4 \] (1 point) 1 real solution, 3 imaginary solutions 2 real solutions, 0 imaginary solutions 2 real solutions, 2 imaginary solutions 4 real solutions, 0 imaginary solutions

Step 1 ：Given the polynomial function \(P(x) = -2x^4 + 2x^2 + 4\)

Step 2 ：This is a quartic polynomial (degree 4). The number of solutions of a polynomial is equal to its degree, so this polynomial has 4 solutions. These solutions can be real or complex (imaginary).

Step 3 ：To simplify the process, we can rewrite the polynomial in terms of \(y = x^2\). Then the polynomial becomes \(P(y) = -2y^2 + 2y + 4\)

Step 4 ：This is now a quadratic equation, and we can use the discriminant of a quadratic equation to determine the nature of its roots. The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c\) is given by \(D = b^2 - 4ac\)

Step 5 ：For the equation \(-2y^2 + 2y + 4\), \(a = -2\), \(b = 2\), and \(c = 4\). So, \(D = (2)^2 - 4*(-2)*4 = 4 + 32 = 36\)

Step 6 ：Since \(D > 0\), the quadratic equation has two distinct real roots.

Step 7 ：However, remember that we substituted \(y = x^2\). So, each real root of the equation in y corresponds to two real roots of the original equation in x (since \(x^2 = y\) has two solutions for each y, namely \(+\sqrt{y}\) and \(-\sqrt{y}\)).

Step 8 ：Therefore, the original quartic polynomial has 4 real solutions.

Step 9 ：\(\boxed{\text{The polynomial has 4 real solutions and 0 imaginary solutions.}}\)