answer:From ( Q(0) = 16 ), we know that ( m = 16 ). From ( Q(1) = 2 ), substituting back into the polynomial gives: [ 1 - 2 + 3 + k + 16 = 2 ] [ k + 18 = 2 ] [ k = -16 ] Thus, the polynomial becomes: [ Q(x) = x^4 - 2x^3 + 3x^2 - 16x + 16 ] To solve: A: [ Q(-2) = (-2)^4 - 2(-2)^3 + 3(-2)^2 - 16(-2) + 16 = 16 - 16 + 12 + 32 + 16 = 60 ] B: Sum of absolute values of coefficients ( = 1 + 2 + 3 + 16 + 16 = 38 ) C: The product of the zeros of ( Q ) is given by the constant term, since the coefficient of ( x^4 ) is 1. Thus, product = ( 16 ). D: [ Q(2) = 2^4 - 2(2)^3 + 3(2)^2 - 16(2) + 16 = 16 - 16 + 12 - 32 + 16 = -4 ] E: Sum of all zeros of ( Q ) (let zeros be ( r_1, r_2, r_3, r_4 )): Using ( -b/a = r_1 + r_2 + r_3 + r_4 ) where ( a = 1 ) and ( b = -2 ), the sum of all zeros is ( 2 ). Thus, the smallest value is: [ boxed{D, text{ which is } -4.} ]

answer:To calculate the total number of wheels in the garage, we need to know the number of wheels each type of vehicle has: - A bicycle has 2 wheels. - A car typically has 4 wheels. - A motorcycle typically has 2 wheels. Now we can calculate the total number of wheels for each type of vehicle: - Bicycles: 20 bicycles * 2 wheels/bicycle = 40 wheels - Cars: 10 cars * 4 wheels/car = 40 wheels - Motorcycles: 5 motorcycles * 2 wheels/motorcycle = 10 wheels Adding these together gives us the total number of wheels in the garage: 40 wheels (bicycles) + 40 wheels (cars) + 10 wheels (motorcycles) = 90 wheels So, there are boxed{90} wheels in the garage at Connor's house.

answer:Because the polynomial has rational coefficients, the radical conjugate of each of the four roots must also be roots. Thus, the pairs of roots will be: - 1 - sqrt{3} and 1 + sqrt{3}, - 3 + sqrt{8} and 3 - sqrt{8}, - 10 - 3sqrt{2} and 10 + 3sqrt{2}, - -sqrt{5} and sqrt{5}. This results in 4 times 2 = 8 roots in total. Constructing the minimal polynomial for each pair: - For 1 pm sqrt{3}, the polynomial is (x - 1)^2 - 3 = x^2 - 2x - 2. - For 3 pm sqrt{8}, the polynomial is (x - 3)^2 - 8 = x^2 - 6x + 1. - For 10 pm 3sqrt{2}, the polynomial is (x - 10)^2 - 18 = x^2 - 20x + 82. - For -sqrt{5} and sqrt{5}, the polynomial is x^2 - 5. Each quadratic contributes two roots, thus a single polynomial encompassing all these roots must be of degree at least 8. Conclusion: A polynomial of degree 8 could suffice if it correctly integrates all the roots from the quadrics mentioned. Given the individual polynomial requirements, the smallest possible degree of a polynomial containing all these roots is boxed{8}.

answer:Let the common ratio of the geometric sequence {a_n} be q. Given S_3 = a_1 + 3a_2, and a_4 = 8, we have: begin{cases} a_{1}+a_{1}q+a_{1}q^{2}=a_{1}+3a_{1}q a_{1}q^{3}=8end{cases} Solving this system of equations, we get: begin{cases} q=2 a_{1}=1end{cases} Therefore, the correct choice is: boxed{A}. To solve this problem, we set up a system of equations based on the properties of a geometric sequence and the given conditions. Solving this system yields the result. This problem tests the understanding of the sum formula and the general term formula of a geometric sequence, involving solving systems of equations, and is considered a medium-level question.