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)=−2x4+2x2+4 (1 point) Responses 4 real solutions, 0 imaginary solutions 4 real solutions, 0 imaginary solutions 2 real solutions, 2 imaginary solutions 2 real solutions, 2 imaginary solutions 2 real solutions, 0 imaginary solutions 2 real solutions, 0 imaginary solutions 1 real solution, 3 imaginary solutions
To determine the number and type of solutions for the polynomial function \( P(x) = -2x^4 + 2x^2 + 4 \), we need to look at the degree of the polynomial and the sign of the leading coefficient. The polynomial is a quartic (degree four) with a negative leading coefficient.
The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) will have exactly \( n \) roots or zeros, when we consider multiplicity and complex roots (real and imaginary solutions). Therefore, this quartic polynomial should have four roots in total.
The possible combinations of real and imaginary roots for a quartic polynomial could be:
- 4 real and 0 imaginary
- 2 real and 2 imaginary
- 0 real and 4 imaginary (counting conjugate pairs of complex roots as two imaginary roots)
To figure out how many real roots the polynomial has, we can look for sign changes in the value of the function as \( x \) changes from negative to positive infinity. This can give us an indication of how many times the graph crosses the x-axis, which corresponds to real roots.
Given the negative leading coefficient, the end behavior of the polynomial will be as follows:
- As \( x \) approaches positive infinity, \( P(x) \) approaches negative infinity.
- As \( x \) approaches negative infinity, \( P(x) \) also approaches negative infinity.
This means that the graph is 'downward' opening on both sides, and since it is a quartic function, it must change direction at least once (it will have at least two turning points). This behavior does not necessarily tell us the exact number of real roots, but it does suggest that there could be either 0, 2, or 4 real roots. The intermediate value theorem can help identify if there are any sign changes that indicate the presence of real roots.
To check for real roots, let's evaluate the function at \( x = 0 \):
\( P(0) = -2(0)^4 + 2(0)^2 + 4 = 4 \)
This result indicates that there is no root at \( x = 0 \). However, it does not preclude the existence of other real roots. Because we don't have a simple way to factor this polynomial or find the roots algebraically, we would typically use other techniques or numerical methods to determine the exact roots.
For a polynomial with real coefficients, complex roots always come in conjugate pairs. Since all the coefficients in this polynomial are real, if there are imaginary roots, there will be an even number of them.
Given this information and without additional tools like graphing calculators or numerical methods, we cannot definitively conclude the exact number and type of solutions, but we can narrow it down to the following possibilities:
- 4 real solutions, 0 imaginary solutions
- 2 real solutions, 2 imaginary solutions (which will be a complex conjugate pair)
- 0 real solutions, 4 imaginary solutions (two complex conjugate pairs)
Without further information, the option of 1 real solution and 3 imaginary solutions can be ruled out because the number of imaginary solutions must be even.
In typical classroom or examination settings, you would either use graphing technology or specific problem-solving techniques provided by your curriculum to further narrow down and find the exact number and type of solutions for the polynomial. However, based on the information provided, we can't give a definitive answer; we simply know the possible scenarios outlined above.