Question

Math picture attached

Answer 1

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

Answer 2

Explanation:

the zero must be complimentary so another zero must be at

c) -3-2i

f(x)=2x^3+2x^2-19x+20

f(-4)=0

f(x)=2x^3+2x^2-19x+20

=(x+4)(2x^2-6x+5)

solving 2x^2-6x+5=0

the zeros are

a) -4, 3/2 +- 1/2i

using the quadratic formula [−b±√(b2−4ac)] / 2a, the zeros for f(x)=11x^2+x+5 are d) -1/22 +- √(219)/22

f(x)=x4-8x3+20x2-32x+64

f(4)=0

f(-4)<>0

so the zeros are c) 4, +-2i