Answer:
Step-by-step explanation:
Since imaginary roots must be in pair and they must be complimentary, if -3 + 2i is a zero, -3 - 2i must be a zero also.
Given f(x) = 2x^3 + 2x^2 - 19x + 20
f(-4) = 2(-4)^3 + 2(-4)^2 - 19(-4) + 20 = 0
So one of the zeros is at -4.
Using long division, divide f(x) by (x + 4)
f(x) = 2x^3 + 2x^2 - 19x + 20
= (x + 4)((2x^2 - 6x + 5)
Find zeroes for 2x^2 - 6x + 5 = 0 with the quadratic formula,
the zeroes for f(x) are at -4, 3/2 +/- 1/2i
f(x) = 11x^2 + x + 5
So a = 11, b = 1, c = 5
The term, b^2 - 4ac = 1^2 -4(11)(5)
= 1 - 220
= -219
The term, -b/2a = -1/22
Combining the above terms, the zeroes are
-1/22 +/- sqrt(219)/22 i
f(x) = x^4 - 8x^3 + 20x^2 - 32x + 64
f(4) = (4)^4 - 8(4)^3 + 20(4)^2 - 32x + 64 = 0
f(-4) = (-4)^4 - 8(-4)^3 + 20(-4)^2 - 32(-4) + 64 = 256 + 512 - 320 + 128 + 64 <>0
So the only choice left is 4, +/-2i
