Question

How do you find the zeroes for `3x^4 - 4x^3 + 4x^2 - 4x + 1`?

Answer 1

The zeros are `x=pm i, 1/3, 1`

Explanation:

The Rational Roots Theorem implies that if `f(x)=3x^4-4x^3+4x^2-4x+1` has any rational roots, they must be `pm 1` or `pm 1/3` (the numerator must divide the constant term and the denominator must divide the coefficient of the highest power).

If you check `1/3` and `1`, you'll see that they work (you can graph the function to guess that they work): `3*(1/3)^4-4*(1/3)^3+4*(1/3)^2-4*(1/3)+1=1/27-4/27+12/27-36/27+27/27=0` and `3-4+4-4+1=0`.

Doing long division (or synthetic division) leads to `f(x)=3x^4-4x^3+4x^2-4x+1=(x-1)(3x^3-x^2+3x-1)=(x-1)(3x-1)(x^2+1)=(x-1)(3x-1)(x-i)(x+i)`

Answer 2

Use the rational zeros theorem (or the sum of the coefficients "shortcut") to find that `1` is a zero, divide by `x-1` and factor.

Explanation:

`3x^4 - 4x^3 + 4x^2 - 4x + 1 = 0` at `x=1` instead of test ing the other possible rational zeros, divide first:

`(3x^4 - 4x^3 + 4x^2 - 4x + 1)/(x-1) = 3x^3-x^2+3x-1`

So,

`3x^4 - 4x^3 + 4x^2 - 4x + 1 = (x-1)[underbrace(3x^3-x^2)+underbrace(3x-1)]`

`= (x-1)[x^2(3x-1)+1(3x-1)]`

`= (x-1)[(x^2+1)(3x-1)]`

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

The real zeros are `1` and `1/3` and imaginary zeros are `+-sqrt(-1)`