Give the zeros of the polynomial: f(x)= x^4 - 3x^3 - 3x^2 + 11x - 6

Answer 1

P = {factors of -6} = {+-1, +-2, +-3, +-6}

Q = {factors of 1} = {+-1}

P/Q= P are the only possible rational solutions

x=1 is a solution because 1^4 - 3*1^3 - 3 * 1^2 + 11*1 - 6 = 1 - 3 - 3 + 11 - 6 = -5 + 11 - 6 = 6-6 =0

Synthetic division says:

1 | 1 -3 -3 11 -6

1 -2 -5 6

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1 -2 -5 6 0

The resulting cubic polynomial is x^3 - 2x^2 - 5x + 6, which has the same set of possible rational solutions.

Again, x=1 is a solution because 1^3 - 2*1^2 - 5*1 + 6 = 1 - 2 - 5 + 6 = -1 - 5 + 6 = -6 + 6 = 0

Synthetic division says:

1 | 1 -2 -5 6

1 -1 -6

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1 -1 -6 0

the resulting quadratic is x^2 - x - 6 = 0

( x - 3)( x + 2 ) = 0

x=3 and x=-2

the complete solution set is x={1,1,-2,3}