Answer: The factor form of the function F(x) is [tex](x+1)(x-2)(x^2+2)[/tex], where [tex](x^2+2)[/tex] gives two complex roots.
Explanation:
The given function is,
[tex]F(x)=x^4-x^3-2x-4[/tex]
According to the rational root theorem 1 and -1 are the possible roots of each polynomial.
Put x=-1
[tex]F(-1)=(-1)^4-(-1)^3-2(-1)-4=0[/tex]
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.
[tex]F(x)=x^4-x^3-2x-4=(x+1)(x^3-2x^2+2x-4)[/tex]
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).
[tex]F(x)=(x+1)(x-2)(x^2+2)[/tex]
Therefore the roots of F(x) are,
[tex]-1,2,\pm i\sqrt{2}[/tex]
[tex]F(x)=(x+1)(x-2)(x+i\sqrt{2})(x-i \sqrt{2} )[/tex]
Where [tex](x+1)(x-2)[/tex] factors with real root and [tex](x+i\sqrt{2})(x-i \sqrt{2} )[/tex] are factors with complex roots.
