# What is the number of real roots of the polynomial?

What is the number of real roots of the polynomial $3s^{3}+10s^{2}+14s+8$?

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What have you tried? Do you have any work to show us? –  anorton Dec 10 '12 at 15:40
You'll get more joy if you show some attempt at answering the question for yourself and indicating where you are stuck! –  Epictetus Dec 10 '12 at 15:45
One.$\mbox{}\mbox{}$. –  Dirk Dec 10 '12 at 15:46
Such questions get quicker (i.e. almost immediate) answers if you ask Wolfram Alpha instead of your fellow human beings. –  MvG Dec 10 '12 at 16:06

As polynomial of odd degree, it has at least one real root.

Between any two real roots of $f$ there is a real root of $f'(x)=9s^2+20s+14$. Since the roots of this are not real (namely $\frac{-20\pm\sqrt{-484}}{18}$), there are no additional real roots.

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Hint:

$3s^3+10s^2+14s+8=(3s^3+6s^2+6s)+(4s^2+8s+8)=(3s+4)(s^2+2s+2)$

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Let $f(s) = 3s^3 + 10s^2 +14s + 8$. Since the degree of $f$ is $3$ (being odd), you know that there is at least one real root.

Then note that $f'(s) = 9s^2 + 20s + 14$. If you have two real roots, then there would (by the Mean Value Theorem) be some $c$ such that $f'(c) = 0$. But you might be able to show that the equation $f'(s) = 0$ has no real solution.

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Why the downvote? –  Thomas Dec 10 '12 at 15:52

Your polynomial factors: $(3s+4)(s^2+2s+2)$. Also $s^2+2s+2=(s+1)^2+1>0$ for all $s$.

So there is only one real root: -4/3.

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Let $f(s)=3s^3+10s^2+14s+8,$ then $$f^\prime (s)=9s^2+20s+14=(3s)^2+2.3s.(10/3)+(10/3)^2+(14-100/9)$$ So $f^\prime(s)=(3s+10/3)^2+26/9\geq 26/9>0$, hence $f$ is strictly increasing and has atmost one root. Now $f(-2)<0$ and $f(0)>0$ and so $f(s)$ has exactly one real root between $0$ and $2$.

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To throw another answer into the pot: The discriminant of the given polynomial is $-400$. But the discriminant of any polynomial with all real roots must be nonnegative.

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At most there is $n$ roots for a $n$-degree polynomial. That is it! For your question you can use the Cardan method. See http://en.wikipedia.org/wiki/Cubic_function

The solution of a third-order algebraic equation of the form $x³+Ax²+Bx+C=0$. By setting $x=((-A)/3)+w$, we have: $w³+Pw+Q=0$, where $P=((-A²)/3)+B$ and $Q=((2A³)/(27))-((AB)/3)+C$. If we set $Δ=4P³+27Q²$, then if $Δ>0$, there is a unique negative real solution:

$x=-(A/3)+(-(Q/2)+√(((Q²)/2)+((P³)/(27))))^{(1/3)}+(-(Q/2)-√(((Q²)/2)+((P³)/(27))))^{(1/3)}$,

and if $Δ<0$, there are three real solutions:

$x₁=-(A/3)+2√(-(P/3))sin((θ/3))$

$x₂=-(A/3)+2√(-(P/3))sin(((θ+2π)/3))$

$x₃=-(A/3)+2√(-(P/3))sin(((θ+4π)/3))$

where

$θ=arcsin(√(((-27Q²)/(4P³))))∈[0,π]$.

The case $Δ=0$ corresponds to a measure-zero set of parameters. Therefore, by a slight perturbation of parameters, without changing the behavior of the system, a system belonging to one of the two cases is obtained.

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You can calculate the descriminant. If it is negative then there is three real roots. If it is positive, then there is one real root and two complex roots. –  user39552 Dec 10 '12 at 16:22
@Riemann: I think that you mean: If it is positive then there are three distinct real roots.. –  Thomas Dec 10 '12 at 16:30
No, I mean the negative case. See the Cardan method. –  user39552 Dec 10 '12 at 16:32
@Riemann: Ok, there is also another notion of a discriminant: en.wikipedia.org/wiki/Discriminant –  Thomas Dec 10 '12 at 16:41