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

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Answer 1 · 2016-03-25T17:04:07

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

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.

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