How to Determine if a Polynomial Has Six Nonreal Zeros?

[darshanpatel]

Homework Statement

Show that the polynomial function:

P(x)=x⁶+2x⁴+3x²+4 has six nonreal zeros

Homework Equations

-none-

The Attempt at a Solution

I tried using synthetic division with all the possible values it could have(p/q) but none of them worked. I was just wondering what it meant by showing it had six nonreal zeros. Is what I did the right work to show that, or no?

My answer: Because all the possible values(±1, ±2, ±4) did not work, it doesn't have any real solutions.

Would that be the correct way to answer the question?

 

[SteamKing]

What you say is true, but you can dress it up by quoting the rational root theorem and appealing to the rule of signs.

 

[Curious3141]

Homework Statement

Show that the polynomial function:

P(x)=x⁶+2x⁴+3x²+4 has six nonreal zeros

Homework Equations

-none-

The Attempt at a Solution

I tried using synthetic division with all the possible values it could have(p/q) but none of them worked. I was just wondering what it meant by showing it had six nonreal zeros. Is what I did the right work to show that, or no?

My answer: Because all the possible values(±1, ±2, ±4) did not work, it doesn't have any real solutions.

Would that be the correct way to answer the question?

What you did (rational root theorem test) only tells you there are no *rational* roots. Doesn't tell you anything about the existence of real, irrational roots.

What you need to do is observe that the substitution $y = x^2$ converts this into a cubic. The original sextic's roots are the square roots of the roots of the cubic.

Sketch the curve of the cubic. Find out if there are any stationary points. Try to estimate (or simply establish bounds on) the only real root of that cubic.

Can you now draw your conclusions?

The other way is to compute a cubic discriminant, but this is tedious, and doesn't give you any more information than the simple curve-sketching approach.

 

[darshanpatel]

What you say is true, but you can dress it up by quoting the rational root theorem and appealing to the rule of signs.

I talked about the rational root theorem but what is the rule of signs?

 

[darshanpatel]

What you did (rational root theorem test) only tells you there are no *rational* roots. Doesn't tell you anything about the existence of real, irrational roots.

What you need to do is observe that the substitution $y = x^2$ converts this into a cubic. The original sextic's roots are the square roots of the roots of the cubic.

Sketch the curve of the cubic. Find out if there are any stationary points. Try to estimate (or simply establish bounds on) the only real root of that cubic.

Can you now draw your conclusions?

The other way is to compute a cubic discriminant, but this is tedious, and doesn't give you any more information than the simple curve-sketching approach.

I have no idea what you are talking about because we have not learned bounds and this section is not about graphing or substituting y=x^2 in

 

[Dick]

I have no idea what you are talking about because we have not learned bounds and this section is not about graphing or substituting y=x^2 in

Do you know every 6th degree polynomial has six roots? That's the fundamental theorem of algebra. If you can show it has no real roots, then you are done. That should be pretty easy if you can show P(x)>0 for any real value of x. Which is almost obvious.

 

[SteamKing]

The rule of signs is also known as Descartes Rule of Signs:

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs

It's one of the big theorems you need to know when looking for the roots of polynomials.

 

[Curious3141]

Homework Statement

Show that the polynomial function:

P(x)=x⁶+2x⁴+3x²+4 has six nonreal zeros

Homework Equations

-none-

The Attempt at a Solution

I tried using synthetic division with all the possible values it could have(p/q) but none of them worked. I was just wondering what it meant by showing it had six nonreal zeros. Is what I did the right work to show that, or no?

My answer: Because all the possible values(±1, ±2, ±4) did not work, it doesn't have any real solutions.

Would that be the correct way to answer the question?

Steamking's suggestion on Descartes' Rule of Signs gives you the most rapid solution. Just figure out the maximum number of positive and negative real roots the equation can have (it's zero in each case). Then just quote the Fundamental Theorem of Algebra to affirm the equation has 6 roots, and therefore they're all nonreal.

 

[HallsofIvy]

You don't really need the full "Rule of Signs". It is sufficient to observe that, since an even power of a real number is non-negative, for any real x, x^6+ 2x^4+ 3x^2+ 4 is greater than or equal to 4.

Corrected thanks to SammyS.

 

[SammyS]

You don't really need the full "Rule of Signs". It is sufficient to observe that, since an even power of a real number is non-negative, for any real x, x^6+ 2x^4+ 3x^3+ 4 is greater than or equal to 4.

Typo. That should be \ x^6+ 2x^4+ 3x^2+ 4\ge 4\ .