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How do you factor the expression x squared plus three?
x^2 + 3 = x^2 - (-3) = x^2 - 3i^2 = (x - i√3)(x + i√3)
What is 3i to the 5th power?
(3i)5 = 35 x i5 = 243 x i2 x i2 x i = 243 x (-1) x (-1) x i = 243i
What is the cube root for 27?
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 :)
Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
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
What is a polynomial with a degree of 3 and zeros of -3 and 3-2i?
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)
