#1**+2 **

\(f(x) = x^4 - x^3 - 2x - 4 \\ \text{I like to quickly check for rational roots before getting fancy}\\ \text{possible roots here are }x = \pm 1, \pm 2, \pm 4 \\ \text{and a quick check shows that }x=-1,~2 \text{ are actual roots}\\ \text{we can then do the polynomial division to find that}\\ f(x) = (x+1)(x-2)(x^2+2)\\ \text{and complete the factoring as}\\ f(x) = (x+1)(x-2)(x+i\sqrt{2})(x - i \sqrt{2})\)

Another way of approaching this is to see if you get lucky when factoring pieces of it.

\(x^4 - x^3-2x-4 = \\ (x^4-4)-(x^3+2x) = \\ (x^2-2)(x^2+2)-x(x^2+2) = \\ (x^2 +2)(x^2 - x-2) =\\ (x^2+2)(x-2)(x+1) = \\ (x+i\sqrt{2})(x-i\sqrt{2})(x-2)(x+1)\\ \text{which is the same as the first answer with the factors listed in different orders}\)

The second method is certainly faster if you get lucky.

Rom Nov 1, 2018

#1**+2 **

Best Answer

\(f(x) = x^4 - x^3 - 2x - 4 \\ \text{I like to quickly check for rational roots before getting fancy}\\ \text{possible roots here are }x = \pm 1, \pm 2, \pm 4 \\ \text{and a quick check shows that }x=-1,~2 \text{ are actual roots}\\ \text{we can then do the polynomial division to find that}\\ f(x) = (x+1)(x-2)(x^2+2)\\ \text{and complete the factoring as}\\ f(x) = (x+1)(x-2)(x+i\sqrt{2})(x - i \sqrt{2})\)

Another way of approaching this is to see if you get lucky when factoring pieces of it.

\(x^4 - x^3-2x-4 = \\ (x^4-4)-(x^3+2x) = \\ (x^2-2)(x^2+2)-x(x^2+2) = \\ (x^2 +2)(x^2 - x-2) =\\ (x^2+2)(x-2)(x+1) = \\ (x+i\sqrt{2})(x-i\sqrt{2})(x-2)(x+1)\\ \text{which is the same as the first answer with the factors listed in different orders}\)

The second method is certainly faster if you get lucky.

Rom Nov 1, 2018