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

Mar 25, 2016

The zeros are $1 , - 1 , 2 \mathmr{and} 2$.

#### Explanation:

The sum of the coefficients in P(x) is 0, 1 is a zero. The sum of the coefficients in $P \left(- x\right)$ 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.