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

1 Answer
Mar 25, 2016

Answer:

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.