Finally, what then would the zeros the a third degree polynomial through no actual 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
request Sep 18 "14 at 4:01
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Why can you not have actually $3$ facility zeros through a $3$rd level polynomial, however it is possible with a $4$th degree polynomial?
Observation: friend can, if the coefficients space themselves complex.
You are watching: A polynomial function with real coefficients has real zeros
Hint: strange exponents keep the sign, and also even ones suppress it. Notification the evident difference between the graphic of cubic and quartic $($or also quadratic$)$ attributes as $x\to\pm\infty$. It is clean from their plots that for every even-order polynomial there is a big-enough cost-free term $($positive or negative$)$ such that the graphic will certainly no much longer intersect the horizontal axis, meaning that ours poly-nomial will have no roots. Yet this walk not use to odd-order polynomials, which always have at least one actual root, due to the fact that they always span from $-\infty$ come $+\infty$, or vice-versa.
reply Sep 18 "14 in ~ 4:26
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If you have a polynomial with genuine coefficients, then complicated roots always come in conjugate pairs. It is however altogether possible that you could a construct a cubic polynomial with three complicated roots -- simply take $(x-z_1)(x-z_2)(x-z_3)$ for any complicated $z_1,z_2,z_3$. However you will discover that once you increase this polynomial out, the coefficients will not it is in real, uneven you choose 3 actual roots, or two complex conjugate root and also one genuine root.
The reason why any polynomial with real coefficients the odd degree (including cubics) must have actually at the very least one genuine root is because the highest possible power term dominates once the change $x$ it s okay large. Suspect the coefficient that the greatest term $x^n$ is 1, because $(-x)^n=-x^n$ once $n$ is odd, then when $x\to\infty$ and $-\infty$, then her polynomial will often tend to $\infty$ and also $-\infty$ respectively. What in in between it have to cross the $x$ axis through the intermediate value theorem, providing us a actual root.
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The reason why facility roots come in conjugate pairs, is because facility conjugation respects enhancement and multiplication. Much more precisely, if $z=x+iy$ and also $w=a+bi$, and also $\overlinez$ to represent the complicated conjugate that $z$, then:
\beginalign*\overlinez+w&=\overlinex+iy+a+bi \\&=\overline(x+a)+(y+b)i \\&=(x+a)-(y+b)i \\&=(x-yi)+(a-bi) \\&=\overlinez+\overlinew\endalign*
and\beginalign*\overlinezw&=\overline(x+iy)(a+bi)\\&=\overline(xa-yb)+(xb+ya)i \\&=(xa-yb)-(xb+ya)i \\&=(x-yi)(a-bi) \\&=\overlinez\cdot\overlinew\endalign*
Repeated applications of the 2nd identity offers $\overlinez^n=(\overlinez)^n$. So suppose we recognize that $x=w$ is a root of a polynomial
where $a_1,\ldots,a_n$ are real numbers. Then
\beginalign*p(\overlinew)&=a_n\overlinew^n+a_n-1\overlinew^n-1+\cdots+a_1\overlinew+a_0 \\&= a_n\overlinew^n+a_n-1\overlinew^n-1+\cdots+a_1\overlinew+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) \\&= 0\endalign*
meaning the $\overlinew$, the facility conjugate was likewise a root, and also therefore complicated roots come in conjugate pairs. Note that the was crucial that the $a_i$ were all real, because then they room equal to their complex conjugate.