We can easily see that (x+-1) are not the factors of f(x).
Now, let us observe that, the leading co-eff. of f(x) is 1, and has
factor 1, and, the const. term is 40, having factors,
1,2,4,5,8,10,20,40.
Therefore, by the Rational Root Theorem, the probable factors can
be worked out :-
1x+-1,1x+-2,1x+-4,x+-5,x+-8,x+-10,+-20, x+-40
Of these, x+-1 have already been considered as non-factors.
As for, x-2, we have, f(2)=8-28+4+40ne0. It is not a factor.
For, x+2, f(-2)=-8-28-4+40=0 :. (x+2) is a factor.
Now to factorise f(x) completely, the Long / Synthetic Division can
be used. Instead, we proceed as under :-
f(x)=x^3-7x^2+2x+40
=ul(x^3+2x^2)-ul(9x^2-18x)+ul(20x+40)
=x^2(x+2)-9x(x+2)+20(x+2)
=(x+2)(x^2-9x+20)
=(x+2){ul(x^2-5x)-ul(4x+20)}........[5xx4=20, 5+4=9]
=(x+2){x(x-5)-4(x-5)}
=(x+2)(x-5)(x-4).
Thus, all the rational zeroes of f(x)=0are, -2, 4, and, 5.
