find all roots real and complex of fx x4 x3 x2 3x 6 0 fi

Solution

f(x) = x^4 + x^3 + x^2 + 3x -6

To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction pq, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient (1) is 1 .The factors of the constant term (-6) are 1, 2 ,3 ,6 . Then the Rational Roots Tests yields the following possible solutions:

+/- 1/1 , +/- 2/1 , +/- 3/1 , +/- 6/1

To find remaining zeros we use Factor Theorem: Divide P(x) with x1

( x^4 + x^3 + x^2 + 3x -6 )/( x-1) = x^3 + 2x^2 +3x +6

The factor of the leading coefficient (1) is 1 .The factors of the constant term (6) are 1, 2 ,3, 6 . Then the Rational Roots Tests yields the following possible solutions:

+/- 1/1 , +/- 2/1 , +/- 3/1 , +/- 6/1

To find remaining zeros we use Factor Theorem. Divide P(x) with x+2

(x^3 + 2x^2 +3x +6)/( x +2) = x^2 +3

x^2 +3 =0----> x = isqrt3, -isqrt3

S, the roots of polynomial : x= 1 ,-2 , i*sqrt3 , -isqrt3