A real polynomial P(x) is the product of real linear and real quadratic factors. Can you that if a real quadratic factor has one zero, a+bi, then the other zero must be the complex conjugate, a-bi?

Hint: Let the quadratic be alphax^2 + betax + gamma and use the properties of complex conjugates

2 Answers
Feb 23, 2017

See below.

Explanation:

P(a+ib)="Re"(P(a+ib))+i "Im"(P(a+ib))

where "Re"(cdot)is the real component of P(a+ib) and
"Im"(cdot) is the real part of the imaginary component. Now

P(a+ib)=0->"Re"(P(a+ib))=0 and "Im"(P(a+ib))=0

but

"Re"(z) = (z+bar z)/2 where bar z is the conjugate of z and
"Im"(z)=(z-bar z)/(2i) so now if P has real coefficients, P(x) = bar P(x) and bar(P(x))=P(bar x) and also

"Re"(P(a+ib)) = (P(a+ib)+P(a-ib))/2

and

"Im"(P(a+ib)) = (P(a+ib)-P(a-ib))/(2i)

So if P(a+ib)=0 then

{(P(a+ib)+P(a-ib)=0),(P(a+ib)-P(a-ib)=0):}

concluding P(a-ib)=0

Another version.

If P(a+ib)=0 then P(x)=Q_1(x)(x-(a+ib)) but P(x) is a real polynomial. That means all its coefficients real. So P(x) must have another root x_0 such that (x-x_0)(x-(a+ib))=x^2+alphax+beta with alpha, beta real. Or

x^2-(x_0+a+ib)x+x_0(a+ib) = x^2+alpha x+beta

or

{(-(x_0+a+ib)=alpha),(x_0(a+ib)=beta):}

The feasible solution is

x_0=a-ib

because

-(a-ib+a+ib) = -2a = alpha and also

(a-ib)(a+ib)=a^2+b^2=beta

with both alpha, beta real.

So if P(x) is a real polynomial and P(a+ib)=0 then necessarily P(a-ib)=0 and P(x)=Q_2(x)(x^2+alpha x + beta) with alpha = -2a and beta = a^2+b^2

Feb 24, 2017

Explanation:

If one of the real quadratic factors is:

alphax^2+betax+gamma

then using the quadratic formula, it has zeros:

(-beta+-sqrt(beta^2-4alphagamma))/(2alpha)

If beta^2-4alphagamma >= 0 then the square root is real and the zeros are real, resulting in real linear factors.

Otherwise, beta^2-4alphagamma < 0 so:

sqrt(beta^2-4alphagamma) = (sqrt(4alphagamma-beta^2))i

resulting in a complex conjugate pair of non-real zeros.