The factor of f(x) corresponding to these two roots is (x — a — ib)(x— a + ib), or (x — a)2 + b2 Let f(x) be divided by (x — a)2 + b2; denote the quotient by Q, and the remainder, if any, by Rx+R’; then f(x)= Q{(x — a)2 + b2}+ Rx +R'. In this identity put x =a + ib, then f(x)= 0 by hypothesis; also (x — a)2 + b2=0; hence R(a + ib)+ R'=0. Equating to zero the real and imaginary parts, Ra + R'=0. Rb = 0 and b by hypothesis is not zero, R=0, R'=0. Hence f(x) is exactly divisible by (x — a)2 + b2=0, that is, by (x—a— ib)(x—a+ib); hence x = a — ib is also a root.
577. In the preceding article we have seen that if the equation f(x)=0 has a pair of imaginary roots a \pm ib, then (x — a)2 + b2=0 is a factor of the expression f(x).
Suppose that a+ib, c+id, e+ ig, \ldots are the imaginary roots of the equation f(x)= 0, and that * \phi (x) is the product of the quadratic factors corresponding to these imaginary roots; then
\phi (x) ={(x — a)2 + b2} {(x) is exactly divisible by (x — a)2 + b2=0, that is, by (x—a— ib)(x—a+ib); hence x = a — ib is also a root.
577. In the preceding article we have seen that if the equation f(x)=0 has a pair of imaginary roots a \pm ib, then (x — a)2 + b2=0 is a factor of the expression f(x).
Suppose that a+ib, c+id, e+ ig, \ldots are the imaginary roots of the equation f(x)= 0, and that * \phi (x) is the product of the quadratic factors corresponding to these imaginary roots; then
\phi (x) ={(x — a)2 + b2} {(x — c)2 + d2}{(x — e)2 + g2}\ldots
Now each of these factors is positive for every real value of x; hence \phi (x) is always positive for real values of a.
578. As in Art. 576 we may show that in an equation with rational coefficients, surd roots enter in pairs; that is, if a+ b is a root then a-b is also a root.
Ex. 1. Solve the equation 6x^4 — 13x^3 - 35x^2-x+3=0, having given that one root is 2 —3.
- The Greek letter Phi.
