Question

How do you find all the zeros of `P(x) = x^4 - 4x^3 + 3x^2 + 4x - 4` where 2 is a zero?

Answer 1

The zeros are `1, -1, 2 and 2`.

Explanation:

The sum of the coefficients in P(x) is 0, 1 is a zero. The sum of the coefficients in `P(-x)` is 0, So,`-1` is a zero. Already, 2 is a zero.

Let P(x) = #(x-1)(x+1)(x-2)(bx+c).

comparing coefficients of `x^4` and the constants in the expansion with the ones in the given P(x), , b = 1 and c =`-2`.

So, P(x) = #(x-1)(x+1)(x-2)(x-2). 2 is a double root.