Finally, what then would the zeros of a 3rd degree polynomial with no real zeros be?
This is the video:https://www.khanacademy.org/2175forals.com/algebra2/polynomial_and_rational/fundamental-theorem-of-algebra/v/fundamental-theorem-of-algebra-intro
asked Sep 18 "14 at 4:01
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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.
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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 $x opminfty$. 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.
answered Sep 18 "14 at 4:26
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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 $x oinfty$ 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.
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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 $overlinez$ represents the complex conjugate of $z$, then:
eginalign*overlinez+w&=overlinex+iy+a+bi \&=overline(x+a)+(y+b)i \&=(x+a)-(y+b)i \&=(x-yi)+(a-bi) \&=overlinez+overlinewendalign*
andeginalign*overlinezw&=overline(x+iy)(a+bi)\&=overline(xa-yb)+(xb+ya)i \&=(xa-yb)-(xb+ya)i \&=(x-yi)(a-bi) \&=overlinezcdotoverlinewendalign*
Repeated application of the second identity gives $overlinez^n=(overlinez)^n$. So suppose we know that $x=w$ is a root of a polynomial
where $a_1,ldots,a_n$ are real numbers. Then
eginalign*p(overlinew)&=a_noverlinew^n+a_n-1overlinew^n-1+cdots+a_1overlinew+a_0 \&= a_noverlinew^n+a_n-1overlinew^n-1+cdots+a_1overlinew+a_0 \&= overlinea_nw^n+overlinea_n-1w^n-1+cdots+overlinea_1w+overlinea_0 \&= overlinea_nw^n+a_n-1w^n-1+cdots+a_1w+a_0 \&= overlinep(w) \&= 0endalign*
meaning that $overlinew$, 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.