answer:To prove that there does not exist a polynomial ( P(a,b,c,d) ) such that the fourth degree polynomial ( x^4 + ax^3 + bx^2 + cx + d ) can be written as the product of four first-degree polynomials if and only if ( P(a,b,c,d) geq 0 ), we will proceed as follows: 1. **Express the polynomial in terms of its roots:** Let ( r_1, r_2, r_3, r_4 ) be the roots of the polynomial ( x^4 + ax^3 + bx^2 + cx + d ). By Vieta's formulas, we have: [ begin{aligned} r_1 + r_2 + r_3 + r_4 &= -a, r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 &= b, r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4 &= -c, r_1r_2r_3r_4 &= d. end{aligned} ] 2. **Define the polynomial ( Q ) in terms of the roots:** Let ( Q(r_1, r_2, r_3, r_4) ) be a polynomial such that ( P(a,b,c,d) = Q(r_1, r_2, r_3, r_4) ). The condition is that ( Q(x, y, z, t) geq 0 ) if ( x, y, z, t in mathbb{R} ) and it becomes negative when ( z, t ) are two conjugate non-real complex numbers. 3. **Consider complex conjugate roots:** Let ( x, y, r in mathbb{R} ) and ( z = zeta ), ( t = bar{zeta} ) where ( |zeta| = r^2 ). Then ( Q(x, y, zeta, bar{zeta}) ) becomes a polynomial because ( zeta = r theta ) and ( bar{zeta} = r frac{1}{theta} ) for some ( theta ) on the unit circle. 4. **Analyze the behavior of ( Q ) for complex roots:** Let ( zeta ) vary over complex numbers with norm ( r^2 ). For every such ( zeta ), ( Q(x, y, zeta, bar{zeta}) < 0 ), except when ( zeta in mathbb{R} ), i.e., ( zeta = pm r ). In these two exceptions, we should have ( Q(x, y, pm r, pm r) geq 0 ). 5. **Continuity argument:** According to the continuity of ( Q(x, y, zeta, bar{zeta}) ), ( Q(x, y, pm r, pm r) ) cannot be positive because if it becomes positive, then there is an (epsilon)-neighborhood of ( r ) where it remains positive, which is absurd. Hence, ( H(r) = Q(x, y, r, r) = 0 ) for all real ( r ). 6. **Contradiction:** Since ( H(r) ) vanishes at every ( r ) and all ( x, y ), we have ( Q(i, -i, 1, 1) = 0 ) where ( i^2 = -1 ), which leads to a contradiction. Therefore, there does not exist a polynomial ( P(a, b, c, d) ) such that the fourth degree polynomial ( x^4 + ax^3 + bx^2 + cx + d ) can be written as the product of four first-degree polynomials if and only if ( P(a, b, c, d) geq 0 ). (blacksquare)

answer:To determine the range of the independent variable x for the function y=sqrt{2-x}, we start by considering the domain of the square root function. The expression inside the square root, 2-x, must be greater than or equal to 0 to ensure that y is a real number. This gives us the inequality: [2-x geqslant 0] Solving this inequality for x involves subtracting 2 from both sides and then multiplying by -1 (remembering to flip the inequality sign when multiplying or dividing by a negative number): [-x geqslant -2] [x leqslant 2] This means that for the function y=sqrt{2-x} to have real values, x must be less than or equal to 2. Therefore, the correct choice is: [boxed{B: xleqslant 2}]

answer:In this scenario, each ball can independently be placed in any of the three boxes. 1. For each of the 5 distinguishable balls, there are 3 choices regarding which box it can be placed in. 2. Therefore, the number of ways to distribute the 5 balls across 3 boxes is calculated by raising the number of choices for each ball to the power of the number of balls, i.e., 3^5. 3. The calculation follows: [ 3^5 = 243 ] Thus, there are boxed{243} ways to put 5 distinguishable balls into 3 distinguishable boxes.

answer:To analyze the given options regarding mutually exclusive and non-opposite events when two balls are randomly selected from a pocket containing 2 red balls and 2 black balls, we proceed as follows: **For event A: At least one black ball and both are black balls.** - The event of having "at least one black ball" includes the possibilities of having one or two black balls. - The event of "both are black balls" is a subset of the event "at least one black ball." - Since these two events can occur simultaneously (when both selected balls are black), they are not mutually exclusive. - therefore Event A is incorrect. **For event B: At least one black ball and at least one red ball.** - The event of having "at least one black ball" includes the possibilities of having one or two black balls, with or without red balls. - The event of having "at least one red ball" includes the possibilities of having one or two red balls, with or without black balls. - These two events can occur simultaneously (when one ball is red and one is black), indicating they are not mutually exclusive. - therefore Event B is incorrect. **For event C: Exactly one black ball and exactly two black balls.** - The event of having "exactly one black ball" cannot occur at the same time as having "exactly two black balls." - These events are mutually exclusive because they cannot happen together. - However, it's possible to select two red balls, which means neither of these events would occur, showing they are not opposite events. - therefore Event C is correct and is the answer we are looking for. **For event D: At least one black ball and both are red balls.** - The event of having "at least one black ball" cannot occur at the same time as "both are red balls." - These events are opposite because if one happens, the other cannot. - However, being opposite is not what we are looking for according to the question. - therefore Event D is incorrect. Based on the analysis, the mutually exclusive and non-opposite events are described in option C: Exactly one black ball and exactly two black balls. boxed{text{C}}