Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
No.
The four roots are:1 + 2i, 1 - 2i, 3i and -3i.
4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.
103
2x3 + 4 = 17 2x3 = 13 x3 = 13 / 2 x = (13 / 2)1/3 x = 521/3 / 2
60
13
p(x) = 0 => x4 + 2x3 + x2 + 8x - 12 = 0=> (x + 3)*(x - 1)*(x2 + 4) = 0So the imaginary roots of p(x) are the imaginary roots of x2 + 4 = 0that is x = ±2i
What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).
the roots are 1, - 1/3, - 3 1/2 and 3 1/2