As there is no independent in f(x)=x^4−4x^3−20x^2+48xf(x)=x4−4x3−20x2+48x and hencexx is factor of f(x)f(x) and hence )) is first zero among the zeros of f(x)f(x). Factorizing f(x)f(x)
f(x)=x(x^3-4x^2-20x+48)f(x)=x(x3−4x2−20x+48)
Now identify factors of 4848 i.e. {1,-1,2,-2,3,-3,4,-4,6,-6,...}, who make (x^3-4x^2-20x+48)=0. As x=2, makes f(x)=0, and hence (x-2) is a factor of f(x) and 2 is another zero of f(x).
Factorizing (x^3-4x^2-20x+48)=(x-2)(x^2-2x-24)
Now (x^2-2x-24), can be easily factorized as -24 can be exactly factorized into -6 and 4, and hence factorization can proceed as under.
x^2-2x-24=x^2-6x+4x-24=x(x-6)+4(x-6)=(x+4)(x-6)
Hence complete factors of f(x)=x^4−4x^3−20x^2+48x) are x(x-2)(x+4)(x-6) and zeros of f(x) are {0,2, -4, 6}
