Factor the polynomial function over the complex numbers
f(x)=x^4−x^3−2x−4
f(x)=x4−x3−2x−4I like to quickly check for rational roots before getting fancypossible roots here are x=±1,±2,±4and a quick check shows that x=−1, 2 are actual rootswe can then do the polynomial division to find thatf(x)=(x+1)(x−2)(x2+2)and complete the factoring asf(x)=(x+1)(x−2)(x+i√2)(x−i√2)
Another way of approaching this is to see if you get lucky when factoring pieces of it.
x4−x3−2x−4=(x4−4)−(x3+2x)=(x2−2)(x2+2)−x(x2+2)=(x2+2)(x2−x−2)=(x2+2)(x−2)(x+1)=(x+i√2)(x−i√2)(x−2)(x+1)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.
f(x)=x4−x3−2x−4I like to quickly check for rational roots before getting fancypossible roots here are x=±1,±2,±4and a quick check shows that x=−1, 2 are actual rootswe can then do the polynomial division to find thatf(x)=(x+1)(x−2)(x2+2)and complete the factoring asf(x)=(x+1)(x−2)(x+i√2)(x−i√2)
Another way of approaching this is to see if you get lucky when factoring pieces of it.
x4−x3−2x−4=(x4−4)−(x3+2x)=(x2−2)(x2+2)−x(x2+2)=(x2+2)(x2−x−2)=(x2+2)(x−2)(x+1)=(x+i√2)(x−i√2)(x−2)(x+1)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.