Question

How do you find all zeros of `f(x) = 2x^4 - 5x^3 +3x^2 + 4x - 6`?

Answer 1

`color(red)(x = -1, x = 3/2, x = 1-i, x = 1+i)`

Explanation:

You use the rational root theorem and the quadratic formula to find the roots.

`f(x) = 2x^4-5x^3+3x^2+4x-6`

According to the rational root theorem, the rational roots of `f(x) = 0` must all be of the form `p/q`, with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `2` of `x^4`.

So the only possible rational roots are:

`±1, ±2, ±3, ±6, ±1/2, ±3/2`

By trial and error, we find:

`f(-1) = 2+5+3-4-6 = 0`

`f(3/2) = 2(3/2)^4-5(3/2)^3+3(3/3)^2+4(3/2)-6`

`= 2(81/16)-5(27/8)+3(9/4)+4(3/2)-6 = 81/8-135/8+27/4+12/2-6 = (81-135+54+48-48)/8 = 0`

So `x = -1` and `x = 3/2` are roots of `f(x) = 0`, and `(x+1)` and `(x-3/2)` are factors of `f(x)`.

Divide `f(x)` by `(x+1)(x-3/2) = x^2+5/2x+3/2` to find:

`(2x^4-5x^3+3x^2+4x-6)/( x^2+5/2x+3/2) = 2x^2-4x+4 =2(x^2-2x+2)`

`f(x) = 2(x+1)(x-3/2)(x^2-2x+2) = (x+1)(2x-3)(x^2-2x+2)`

Using the quadratic formula, we find that the roots of `x^2-2x+2 = 0` are:

`x = (-(-2)±sqrt((-2)^2-4×1×2))/(2×1) = (2±sqrt(4-8))/2 =(2±sqrt(-4))/2 = (2±2i)/2 = 1±i`

The zeroes of `f(x) = 2x^4-5x^3+3x^2+4x-6` are `x = -1, x = 3/2, x = 1-i`, and `x = 1+i`.