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
Cesareo R.
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#

George C.
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.

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
2532 views around the world
You can reuse this answer
Creative Commons License