Find all the zeroes of the function.?

Question

h(x)=3x^4+8x^3+9x^2+32x-12

Answer 1

3x^4+8x^3+9x^2+32x-12 = (x+3)(3x-1)(x^2+4) = 0
x+3 = 0, 3x-1 = 0, x^2+4 = 0
The real zeros are x = -3, 1/3
The imaginary zeros are, x = 2i, -2i

Answer 2

By the rational roots theorem, the only possible rational roots are 12, -12, 4, -4, 4/3, -4/3, 3, -3, 1, -1, 1/3, -1/3. Testing all these values, it follows that -3 and 1/3 are roots of the function, and so it can be factored as 3x^4+8x^3+9x^2+32x-12=(x+3) (3x-1) (x^2+4) = (x+3) (3x-1) (x+2i) (x-2i). So, the zeroes are -3, 1/3, 2i, and -2i.

Answer 3

You need to find two rational ones and divide by the corresponding factors to reduce it to a second degree polynomial.

One rational root is -3, so divide by (x+3)

That gives 3x^3 - x^2 + 12x - 4

Another is 1/3 so divide this new answer by (3x - 1)

That gives x^2 + 4

This has two complex roots which you can easily find.

You find the original rational roots by knowing that the top of the rational root is a factor of the constant term 12 and the bottom is a factor of the leading coefficient 3, and some lucky guessing.

Answer 4

-3, 1/3, 2 i, -2 i