In this identity, by equating the coefficients of h, we have
S'@)=0(@— a4 (2) + (@— @'$"(@).
Thus f'(x) contains the factor x— a repeated r—1 times; that is, the equation f'(x)=0 has r —1 roots equal to a.
Similarly we may show that if the equation f(x)= 0 has s roots equal to 0, the equation f'(x)= 0 has s — 1 roots equal to b; and so on.
From the foregoing proof we see that if f(x) contains a factor (x — a)r, then f'(a) contains a factor (x— a)r-1; and thus f(x) and f'(@) have a common factor (r — a)r-1. Therefore if f(a) and f'(x) have no common factor, no factor in f(x) will be repeated; hence the equation f(x)=0 has or has not equal roots, according as f(x) and f'(x) have or have not a common factor involving x.
596. It follows that in order to obtain the equal roots of the equation f(x)= 0, we must first find the highest common factor of f(x) and f'(x), and then placing it equal to zero, solve the resulting equation.
Ex. Solve the equation 2t— 1123+ 4422— 76% +48 = 0, which has equal roots.
Here S(@) = a4 — 1128 + 4492 — 76% + 48, S'(«) = 423 — 334? + 88x — 76;
and by the ordinary rule we find that the highest common factor of f(x) and f'(x) is x — 2; hence (x— 2)2 is a factor of f(x); and
S(@) =(@ — 2)2(a? — 7x + 12) =(x— 2)2(x — 3)(x— 4);
thus the roots are 2, 2, 3, 4.
