Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
What have you tried? Do you have any work to show us? –  anorton Dec 10 '12 at 15:40
2  
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
1  
One.$\mbox{}\mbox{}$. –  Dirk Dec 10 '12 at 15:46
4  
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
add comment

7 Answers

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.

share|improve this answer
add comment

Hint:

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

share|improve this answer
add comment

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.

share|improve this answer
    
Why the downvote? –  Thomas Dec 10 '12 at 15:52
add comment

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.

share|improve this answer
add comment

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 $2$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.