negative 15 is the answer
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x^2 + 3 = x^2 - (-3) = x^2 - 3i^2 = (x - i√3)(x + i√3)
(3i)5 = 35 x i5 = 243 x i2 x i2 x i = 243 x (-1) x (-1) x i = 243i
The cube root of 27 is 3. 27 has three cube roots, a real cube root (3) and two imaginary cube roots. To compute them, start with the following equation and solve for x: x^3 = 27 x^3 - 27 = 0 Factor using the difference of cubes: (x - 3)(x² + 3x + 9) = 0 One root is x = 3. The other two can be found using the quadratic formula on: x² + 3x + 9 = 0 a = 1 b = 3 c = 9 x = [-3 ± √(9 - 4*1*9) ] / 2 x = (-3 ± √(-27) ] / 2 x = (-3 ± 3i√3 ) / 2 x = -3/2 ± (3i√3)/2 Answers: 3 -3/2 + (3i√3)/2 -3/2 - (3i√3)/2 Hope I helped :)
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
The question has a simple answer only if the polynomial has rational coefficients. However, the question does not state that it has rational coefficients so it is not valid to assume that is the case. So, suppose the third root is p. Then the polynomial is (x + 3)*(x - 3 + 2i)*(x - p) = (x2 + 2xi - 9 + 6i)*(x - p) = x3 + (2i - p)x2 - (2pi - 6i + 9)x + (9p + 6pi)