How Do You Solve a Cubic Polynomial with No Rational Roots?

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To solve the cubic polynomial t^3 + 2t^2 + 1 = 0, it is confirmed that there are no rational roots, making traditional methods like synthetic division ineffective. The polynomial is noted to have one irrational root and two complex roots, which complicates hand-solving. The cubic formula is mentioned as a potential solution method, although it is not commonly used by all students. There is uncertainty about whether textbooks would require the use of the cubic formula for such problems. Overall, solving this polynomial by hand is challenging due to the nature of its roots.
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Hello,

I am trying to solve a degree three polynomial, but unfortunately I am stuck

t3+2t2+1=1

This is as far as I can get:
t(t2+2t)+1=0
Where do I go from here?

Thanks!
 
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I can get the solution by using my graphing calculator, but I would like to know how to do it by hand :)
 
Is the original problem t3+2t2+1=1

or t3+2t2+1=0 ??

You've written down two different things.
 
t3+2t2+1=0
sorry :)
 
It looks like this polynomial has one irrational root and two complex roots, so it can't be solved by hand (unless you know the cubic formula).
 
Yes, it cannot be solved using normal methods. There are no rational roots so synthetic division wouldn't work. Are you sure that it can be solved? Does the back of the book have the answer?

I've never used the cubic formula to solve a polynomial, and I doubt a textbook would require you to do that.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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