The function F ( x ) = X5 - 4x4 + 4x3 + 2x2 - 5x +2 .

Identify Rational Zeros :

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation anxn + an  1xn – 1 + ... + a1x + a0 = 0, then p is a factor of a0 and q is a factor if an.

If p/q is a rational zero, then p is a factor of 2 and q is a factor of 1.

The possible values of p are   ± 1, ± 2 .

The possible values for q are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

Make a table for the synthetic division and test possible real zeros.

 p/q 1 - 4 4 2 - 5 2 1 1 - 3 1 3 - 2 0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x4 -3x3 + x2 + 3x - 2 = 0.

Now consider x4 -3x3 + x2 + 3x - 2 = 0 .

If p/q is a rational zero, then p is a factor of -2 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

Make a table for the synthetic division and test possible real zeros.

 p/q 1 - 3 1 3 - 2 1 1 - 2 - 1 2 0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x3 -2x2 - x + 2 = 0.

Now consider x3 -2x2 - x + 2 = 0 .

If p/q is a rational zero, then p is a factor of 2 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2 .

Make a table for the synthetic division and test possible real zeros.

 p/q 1 -2 - 1 2 1 1 -1 - 2 0

Since f(1) = 0, x = 1 is a zero. The depressed polynomial is  x2 - x - 2 = 0.

Now consider x2 - x - 2 = 0 .

x2 - x - 2 = 0

x2 - 2x + x - 2 = 0

x(x-2) +1(x-2) = 0

(x-2)(x+1) = 0

x - 2 = 0 and x +1 = 0

x = 2 and x = -1

So the solutions for the polynomial are -1,1,2 .

So first option is correct .