Question
write a fourth degree polynomial function with real coefficients that has -3,1/5, and 4+i as zeros and the y intercept of (0,5)
Answers
Steve
The other root must be 4-i, so one factor is
(x-(4+i))(x-(4-i)) = ((x-4)^2 + 1^2) = x^2-8x+17
So, now we have
(x+3)(5x-1)(x^2-8x+17)
at x=0, this is (3)(-1)(17) = -51, so
y = -5/51 (x+3)(5x-1)(x^2-8x+17)
= -5/51 (5x^4 - 26x^3 - 30x^2 + 262x - 51)
(x-(4+i))(x-(4-i)) = ((x-4)^2 + 1^2) = x^2-8x+17
So, now we have
(x+3)(5x-1)(x^2-8x+17)
at x=0, this is (3)(-1)(17) = -51, so
y = -5/51 (x+3)(5x-1)(x^2-8x+17)
= -5/51 (5x^4 - 26x^3 - 30x^2 + 262x - 51)