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I was watching the Khan Academy video on the Fundamental Theorem of Algebra when I got confused by something that Sal Khan states. From what I understand, the Theorem says that the complex zeros of a polynomial function always come in pairs because they are conjugates of each other. But exactly what would happen with a 3rd degree polynomial with no real zeros? Why can you not have 3 complex zeros with a 3rd degree polynomial but it is possible with a 4th degree polynomial?

Finally, what then would the zeros of a 3rd degree polynomial with no real zeros be?

This is the video: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/fundamental-theorem-of-algebra/v/fundamental-theorem-of-algebra-intro

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    $\begingroup$ That statement is only true for polynomials with real coefficients, so you need to be careful. $\endgroup$ Sep 18, 2014 at 4:10
  • $\begingroup$ You can have a polynomial with three complex zeroes. Try $(x^2+1)(x-i)$. The point is that if the coefficients are real, then one needs a complex conjugate to get rid of the imaginary parts in the coefficients. $\endgroup$ Sep 18, 2014 at 4:12
  • $\begingroup$ If you have a moment, please check out the video from 4:38 on and you will see what I am confused about. $\endgroup$ Sep 18, 2014 at 4:15
  • $\begingroup$ The video is very sloppy, because shortly after declaring that polynomials can have complex coefficients they repeat the (unqualified) claim that non-real complex roots always come in conjugate pairs. That's just wrong. $\endgroup$ Sep 18, 2014 at 9:41

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If you have a polynomial with real coefficients, then complex roots always come in conjugate pairs. It is however altogether possible that you could a construct a cubic polynomial with three complex roots -- just take $(x-z_1)(x-z_2)(x-z_3)$ for any complex $z_1,z_2,z_3$. However you will find that when you expand this polynomial out, the coefficients will not be real, unless you picked 3 real roots, or two complex conjugate root and one real root.

The reason why any polynomial with real coefficients of odd degree (including cubics) must have at least one real root is because the highest power term dominates when the variable $x$ gets large. Assuming the coefficient of the highest term $x^n$ is 1, since $(-x)^n=-x^n$ when $n$ is odd, then when $x\to\infty$ and $-\infty$, then your polynomial will tend to $\infty$ and $-\infty$ respectively. Somewhere in between it must cross the $x$ axis by the intermediate value theorem, giving us a real root.

The reason why complex roots come in conjugate pairs, is because complex conjugation respects addition and multiplication. More precisely, if $z=x+iy$ and $w=a+bi$, and $\overline{z}$ represents the complex conjugate of $z$, then:

\begin{align*}\overline{z+w}&=\overline{x+iy+a+bi} \\ &=\overline{(x+a)+(y+b)i} \\ &=(x+a)-(y+b)i \\ &=(x-yi)+(a-bi) \\ &=\overline{z}+\overline{w} \end{align*}

and \begin{align*} \overline{zw}&=\overline{(x+iy)(a+bi)}\\ &=\overline{(xa-yb)+(xb+ya)i} \\ &=(xa-yb)-(xb+ya)i \\ &=(x-yi)(a-bi) \\ &=\overline{z}\cdot\overline{w} \end{align*}

Repeated application of the second identity gives $\overline{z^n}=(\overline{z})^n$. So suppose we know that $x=w$ is a root of a polynomial

$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$

where $a_1,\ldots,a_n$ are real numbers. Then

\begin{align*} p(\overline{w})&=a_n\overline{w}^n+a_{n-1}\overline{w}^{n-1}+\cdots+a_1\overline{w}+a_0 \\ &= a_n\overline{w^n}+a_{n-1}\overline{w^{n-1}}+\cdots+a_1\overline{w}+a_0 \\ &= \overline{a_nw^n}+\overline{a_{n-1}w^{n-1}}+\cdots+\overline{a_1w}+\overline{a_0} \\ &= \overline{a_nw^n+a_{n-1}w^{n-1}+\cdots+a_1w+a_0} \\ &= \overline{p(w)} \\ &= 0 \end{align*}

meaning that $\overline{w}$, the complex conjugate was also a root, and therefore complex roots come in conjugate pairs. Note that it was crucial that the $a_i$ were all real, because then they are equal to their complex conjugate.

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Why can you not have $3$ complex zeros with a $3$rd degree polynomial, but it is possible with a $4$th degree polynomial?

Observation: You can, if the coefficients are themselves complex.

Hint: Odd exponents maintain the sign, and even ones suppress it. Notice the obvious difference between the graphics of cubic and quartic $($or even quadratic$)$ functions as $x\to\pm\infty$. It is clear from their plots that for every even-order polynomial there is a big-enough free term $($positive or negative$)$ such that the graphic will no longer intersect the horizontal axis, meaning that our poly-nomial will have no roots. But this does not apply to odd-order polynomials, which always have at least one real root, since they always span from $-\infty$ to $+\infty$, or vice-versa.

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Any degree $3$ polynomial with real coefficients has at least one real zero. In fact any polynomial of odd degree with real coefficients has at least one real zero.

For simplicity consider the polynomial $p(x)=x^3+bx^2+cx+d$. When $x$ is very large negative, $p(x)$ is negative. When $x$ is very large positive, $p(x)$ is positive. So, by the Intermediate Value Theorem, somewhere in between we have $p(x)=0$.

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  • $\begingroup$ But what happens to a 3rd degree polynomial with no real zeros? $\endgroup$ Sep 18, 2014 at 4:06
  • $\begingroup$ @Cherry_Developer There aren't any. $\endgroup$ Sep 18, 2014 at 4:09
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    $\begingroup$ A degree $3$ polynomial with real coefficients always has at least one real zero. Of course if the polynomial has some non-real coefficients, then there may be no real zero. Exanple: $x^3+ix^2+x+i$. $\endgroup$ Sep 18, 2014 at 4:10
  • $\begingroup$ But then, the statement from Khan is not true - the complex roots don't come in pairs. So Khan is clearly talking about polynomials with real coefficients. @AndréNicolas $\endgroup$ Sep 18, 2014 at 4:11
  • $\begingroup$ So a polynomial of the 4th degree with no real zeros has 4 complex zeros. What roots does the polynomial of the 3rd degree have? $\endgroup$ Sep 18, 2014 at 4:14
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It seems to me that either Mr. Khan or you are overlooking (for want of a better word) this well-known result:

Every odd-degree poynomial equation with real coefficients has a real zero.

This implies in particular that there is no such thing as a 3rd degree polynomial equation (with real coefficients) with no real zeros.

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