Question

Factor the polynomial function over the complex numbers?? F(x)= x^4 -x^3-2x-4

Answer 1

Answer: The factor form of the function F(x) is $(x+1)(x-2)(x^2+2)$, where  $(x^2+2)$ gives two complex roots.

Explanation:

The given function is,

$F(x)=x^4-x^3-2x-4$

According to the rational root theorem 1 and -1 are the possible roots of each polynomial.

Put x=-1

$F(-1)=(-1)^4-(-1)^3-2(-1)-4=0$

Since the value of F(x) is 0 at x=-1 therefore the -1 is a root of given polynomial and (x+1) is a factor of F(x).

Use synthetic division method to divide the polynomial by (x+1).

The last row of the synthetic division shows the coefficient of remaining polynomial.

$F(x)=x^4-x^3-2x-4=(x+1)(x^3-2x^2+2x-4)$

At x=2 the value of F(x) is 0, therefore (x-2) is a factor of F(x).

Use synthetic division method to divide the polynomial by (x-2).

$F(x)=(x+1)(x-2)(x^2+2)$

Therefore the roots of F(x) are,

$-1,2,\pm i\sqrt{2}$

$F(x)=(x+1)(x-2)(x+i\sqrt{2})(x-i \sqrt{2} )$

Where $(x+1)(x-2)$ factors with real root and $(x+i\sqrt{2})(x-i \sqrt{2} )$ are factors with complex roots.