Find the zeros of: f(x) = -3x^4+x^3+51x^2-5x-4

Question

Some help would be greatly appreciated and some explanation as well.

Answer 1

f(x)=-(3x^4-x^3-51x^2+5x+4)=-(x+4)(3x^3-13x^2+x+1)=-(x+4)(3x-1)(x^2-4x-1).

The rational factors can be found by trial and error or by sketching the graph and looking for the x intercepts (where the graph cuts the x axis).

To factorise the quadratic we can complete the square: x^2-4x+4-5=(x-2)^2-5=(x-2-sqrt(5))(x-2+sqrt(5)).

Therefore the zeroes are: -4, 1/3, 2-sqrt(5) (=-0.23607), 2+sqrt(5) (=4.23607).