answer:The completely factored form of ( f(x) ) is ( (2x - 3)(3x - 5)(2x - 1) ). Explanation: To factor ( f(x) ), we can use the Rational Root Theorem, which states that any rational root of ( f(x) ) can be expressed as ( frac{p}{q} ), where ( p ) is a divisor of the constant term (-15) and ( q ) is a divisor of the leading coefficient (12). Potential rational roots include: [ pmfrac{1, 3, 5, 15}{1, 2, 3, 4, 6, 12} ] However, we can use Descartes' Rule of Signs to narrow down the search. The function has one sign change, suggesting one or three positive real roots. By considering the derivative, we can find local extrema to help identify these roots. The derivative, ( f'(x) = 36x^2 - 88x + 49 ), has a zero at ( x = frac{22}{3} pm frac{sqrt{43}}{3} ), which are not rational. However, this helps us find the local extrema at ( x approx frac{5}{6} ) and ( x approx frac{3}{2} ). Evaluating ( f(x) ) at these points, we find ( fleft(frac{5}{6}right) approx frac{20}{9} ) and ( fleft(frac{3}{2}right) = 0 ), revealing ( frac{3}{2} ) as a zero with a corresponding factor ( (2x - 3) ). Next, we find another zero near ( frac{3}{2} ) at ( x = frac{5}{3} ), which corresponds to the factor ( (3x - 5) ). The remaining factor, ( (2x - 1) ), is deduced by considering the leading coefficient and constant term. Thus, ( f(x) ) factors into: [ 12x^3 - 44x^2 + 49x - 15 = (2x - 3)(3x - 5)(2x - 1) ] [ text{graph}[12x^3 - 44x^2 + 49x - 15, -0.448, 2.052, -1.26, 2.49] ]

answer:To prove that the hyperbola V(xy-1) is isomorphic to the punctured line IA^1setminus{0}, we can consider the functions f and g: 1. f from V(xy-1) to IA^1, defined by f(x,y) = x. 2. g from IA^1setminus{0} to V(xy-1), defined by g(t) = (t, 1/t). We can apply Lemma 3.6 in Hartshorne: - To show that f is a morphism of varieties, we need to verify that f is a regular function. By the definition of regular functions, this is clear since x is a coordinate function on V(xy-1). - For g to be a morphism, we must show that x after g and y after g are regular functions on IA^1setminus{0}. Here, x after g = t and y after g = 1/t, both of which are locally quotients of polynomials, hence regular. It is crucial to note that bijectivity alone does not guarantee an isomorphism. Both f and its inverse g must be morphisms of varieties. As demonstrated, both f and g satisfy the conditions to be morphisms, establishing the isomorphism.

answer:To find the value of frac{x}{y}, we can use the formula for dividing complex numbers: frac{x}{y} = frac{x overline{y}}{y overline{y}} where overline{y} is the complex conjugate of y. Substituting the given values, we have: frac{x}{y} = frac{(9-5 i)(9-3 i)}{(9+3 i)(9-3 i)} = frac{(9-5 i)(9-3 i)}{9^2-(3 i)^2} = frac{(9-5 i)(9-3 i)}{81-9 i^2} = frac{(9-5 i)(9-3 i)}{81+9} = frac{(9-5 i)(9-3 i)}{90} = frac{81-27 i-45 i+15 i^2}{90} = frac{81-72 i+15 (-1)}{90} = frac{66-72 i}{90} = frac{11}{15}-frac{4}{15}i Therefore, the value of frac{x}{y} is frac{11}{15}-frac{4}{15}i. The answer is frac{11}{15}-frac{4}{15}i

answer:The error lies in the calculation of y_i. The original calculation uses the formula: y_i:=-lnleft(-lnleft(frac{i}{n+1} right)right) This formula should be used instead of the empirical distribution function values used in the provided calculation. By using the correct formula for y_i, the calculated values of alpha and beta will match the original parameters.