POLYNOMIALS WITH REAL NUMBER COEFFICIENTS

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EXAMPLE (finding the complex conjugate zeros of a quadratic function):

• Consider the polynomial $\,P(x) = x^2 - 4x + 13\,,$ which has real number coefficients.
  By the quadratic formula, the zeros of $\,P\,$ (i.e., the solutions of $\,x^2 - 4x + 13 = 0\,$) are: $$ \cssId{s15}{x \ =\ \frac{-(-4) \pm \sqrt{16 - 4(1)(13)}}{2(1)}} \cssId{s16}{\ =\ \frac{4\pm \sqrt{-36}}{2}} \cssId{s17}{\ =\ \frac{4}{2} \pm \frac{\sqrt{-36}}{2}} \cssId{s18}{\ =\ 2 \pm \frac{i\sqrt{36}}{2}} \cssId{s19}{\ =\ 2 \pm \frac{i\cdot 6}{2}} \cssId{s20}{\ =\ 2\pm 3i} $$
• Thus, $\,P(x) = x^2 - 4x + 13\,$ has two zeros: $\,2 + 3i\,$ and $\,2 - 3i\,.$
  These two zeros consist of a complex number together with its conjugate—that is, a complex conjugate pair.
• To verify that $\,2+3i\,$ is a zero of $\,P\,,$ we have: $$ \begin{align} \cssId{s25}{P(2 + 3i)}\ &\cssId{s26}{= (2+3i)^2 - 4(2+3i) + 13}\cr &\cssId{s27}{= (2+3i)(2+3i) - 8 - 12i + 13}\cr &\cssId{s28}{= 2^2 + 6i + 6i + 9i^2 - 8 - 12i + 13}\cr &\cssId{s29}{= 4 + 12i - 9 + 5 - 12i}\cr &\cssId{s30}{= 0} \end{align} $$ Similarly, you should verify that $\,P(2 - 3i) = 0\,.$

WHY MUST NON-REAL ZEROS OF POLYNOMIALS WITH REAL COEFFICIENTS
OCCUR IN COMPLEX CONJUGATE PAIRS?

We'll show that for every quadratic function with real nunber coefficients, non-real zeros must occur in complex conjugate pairs.
The proof here easily generalizes to any polynomial with real number coefficients.

Recall that the complex conjugate of $\,x + yi\,$ (for real numbers $\,x\,$ and $\,y\,$) is $\,x - yi\,$;
that is, to find the complex conjugate, the real part remains the same, and you take the opposite of the imaginary part.
The complex conjugate is frequently denoted with an overbar: $$ \cssId{s39}{\underbrace{\overline{x + yi}}_{\text{the complex conjugate of $\,x+yi\,$}}} \cssId{s40}{= \underbrace{x - yi}_{\text{same real part; opposite imaginary part}}} $$

Here are two properties of the complex conjugate that were proved in an earlier lesson:

• the complex conjugate of a sum is the sum of the complex conjugates: e.g., $\,\overline{a + b} = \overline{\vphantom{b}a} + \overline{b}$
• the complex conjugate of a product is the product of the complex conjugates: e.g., $\overline{ab} = \overline{\vphantom{b}a}\cdot\overline{b}$

PROOF:
Let $\,P(x) := ax^2 + bx + c\,,$ where $\,a\,,$ $\,b\,$ and $\,c\,$ are real numbers, and $\,a\ne 0\,.$
Suppose that $\,w\,$ is a zero of $\,P\,.$
Then: $$ \begin{alignat}{2} \cssId{s50}{P(\bar w)}\ &\cssId{s51}{= a(\bar w)^2 + b\bar w + c} &&\qquad \cssId{s52}{\text{REASON: definition of $P$, function evaluation}}\cr &\cssId{s53}{= a(\overline{w^2}) + b\bar w + c} &&\qquad \cssId{s54}{\text{REASON: } \overline{\vphantom{b}a}\cdot\overline{b} = \overline{ab}}\cr &\cssId{s55}{= \bar a(\overline{w^2}) + \bar b\bar w + \bar c} &&\qquad \cssId{s56}{\text{REASON: for real numbers $x$, $\bar x = x$}}\cr &\cssId{s57}{= \overline{a\,w^2} + \overline{bw} + \bar c} &&\qquad \cssId{s58}{\text{REASON: } \overline{\vphantom{b}a}\cdot\overline{b} = \overline{ab}}\cr &\cssId{s59}{= \overline{aw^2 + bw + c}} &&\qquad \cssId{s60}{\text{REASON: } \overline{\vphantom{b}a} + \overline{b} = \overline{a + b}}\cr &\cssId{s61}{= \overline{P(w)}} &&\qquad \cssId{s62}{\text{REASON: definition of $P$, function evaluation}}\cr &\cssId{s63}{= \bar 0} &&\qquad \cssId{s64}{\text{REASON: definition of zero}}\cr &\cssId{s65}{= 0} &&\qquad \cssId{s66}{\text{REASON: for real numbers $x$, $\bar x = x$}}\cr \end{alignat} $$ Q.E.D.

The Factors Corresponding to the Zeros in a Complex Conjugate Pair
Multiply to give an Irreducible Quadratic

Recall that there is a beautiful relationship between the zeros of a polynomial and its factors:   if $\,c\,$ is a zero, then $\,x-c\,$ is a factor.
In particular:

• the factor corresponding to the zero $\,a + bi\,$ is:   $\,(x - (a + bi)) = x - a - bi$
• the factor corresponding to the zero $\,a - bi\,$ is:   $\,(x - (a - bi)) = x - a + bi$

Multiplying these two different linear factors gives an irreducible quadratic: $$ \begin{align} \cssId{s77}{(x - a - bi)(x - a + bi)}\ &\cssId{s78}{= x(x - a + bi) - a(x - a + bi) - bi(x - a + bi)}\cr &\cssId{s79}{= x^2 - ax + bxi - ax + a^2 - abi - bxi + abi - b^2i^2}\cr &\cssId{s80}{= x^2 - 2ax + (a^2 + b^2)} \end{align} $$ We can check that the resulting quadratic $\,x^2 - 2ax + (a^2 + b^2)\,$ is irreducible by showing that its discriminant is negative: $$ \begin{align} \cssId{s82}{\text{discriminant of $\,x^2 - 2ax + (a^2 + b^2)\,$}} &\cssId{s83}{= (-2a)^2 - 4(1)(a^2 + b^2)}\cr &\cssId{s84}{= 4a^2 - 4a^2 - 4b^2}\cr &\cssId{s85}{= -4b^2}\cr &\cssId{s86}{< 0} \end{align} $$

Now, put everything together:

• Start with a polynomial with real coefficients.
• This polynomial may have both real zeros and non-real zeros.
• The real zeros give linear factors: if $\,c\,$ is a zero, then $\,x - c\,$ is a factor.
• If $\,a+bi\,$ is any non-real zero, then $\,a - bi\,$ must also be a zero.
  The two corresponding linear factors multiply together to give the irreducible quadratic: $\,x^2 - 2ax + (a^2 + b^2)$
• Thus:
  EVERY POLYNOMIAL WITH REAL NUMBER COEFFICIENTS
  CAN BE FACTORED INTO LINEAR FACTORS AND IRREDUCIBLE QUADRATICS

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
solving polynomial equations