Find a monic quartic polynomial f(x) with rational coefficients whose roots include x=2-3 sqrt2 and x = 1 - sqrt3. Give your answer in expanded form.

 Jan 4, 2018

If the roots are \(x=2-\sqrt{3}\) and \(x=1-\sqrt{3}\), then the conjugates must also be roots as well, in order for it to be a polynomial. Therefore, \(x=2+\sqrt{3}\) and \(x=1+\sqrt{3}\) are also roots of this unknown polynomial.


With these roots, we can generate factors of the given polynomial, too.


\(x=2-\sqrt{3}\\ \hspace{1mm}-\left(2-\sqrt{3}\right)\) \(x=2+\sqrt{3}\\ \hspace{1mm}-\left(2+\sqrt{3}\right)\) \(x=1-\sqrt{3}\\ \hspace{1mm}-\left(1-\sqrt{3}\right)\) \(x=1+\sqrt{3}\\ \hspace{1mm}-\left(1+\sqrt{3}\right)\) By subtracting the right hand side of the equation, we can then see what the factors are.
\(x-\left(2-\sqrt{3}\right)=0\) \(x-\left(2+\sqrt{3}\right)=0\) \(x-\left(1-\sqrt{3}\right)=0\) \(x-\left(1+\sqrt{3}\right)=0\)  
\(x-2+\sqrt{3}=0\) \(x-2-\sqrt{3}=0\) \(x-1+\sqrt{3}=0\) \(x-1-\sqrt{3}=0\) Since all of these are set equal to zero, these all must be factors of the original unknown polynomial.


As aforementioned, they are factors of the original equation. If they are factors, then we can multiply them together and figure out what the original polynomial is.




We know that the left hand side is equal to zero since each factor equals zero. 0 multiplied by itself four times still equals zero. Now, expand. I recommend multiplying factors that are conjugates. This might make the process easier. We can multiply in any order, too. I'll expand the bit in red first.


\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\) Make sure to remain attentive. Be sure to multiply every factor.
\(\textcolor{red}{x^2-2x-x\sqrt{3}-2x+4+2\sqrt{3}+x\sqrt{3}-2\sqrt{3}-3}\) Wow! Those are a lot of terms. Let's rearrange the equation, though, to see if any cancelling can occur here. (Spoiler: There is!)
\(\textcolor{red}{x^2-x\sqrt{3}+x\sqrt{3}+2\sqrt{3}-2\sqrt{3}-2x-2x+4-3}\) Look at that! All the radical expression cancel out! 
\(\textcolor{red}{x^2-4x+1}\) Wow, that monstrosity has been simplified quite beautifully. Don't you agree?


Let's do the exact same process with the blue bit. It should also simplify to a quadratic trinomial.


\(\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}\) Ok, time to expand again! We already have an idea of what should occur.
\(\textcolor{blue}{x^2-x-x\sqrt{3}-x+1+\sqrt{3}+x\sqrt{3}-\sqrt{3}-3}\) Let's do that rearranging again!
\(\textcolor{blue}{x^2-x-x-x\sqrt{3}+x\sqrt{3}+\sqrt{3}-\sqrt{3}+1-3}\) Yet again, as expected, the radicals will cancel out.


Now, we must multiply the red and blue parts together to get a quartic polynomial.


\((\textcolor{red}{x^2-4x+1})(\textcolor{blue}{x^2-2x-2})=0\) We have to expand one more time!
\(x^4-2x^3-2x^2-4x^3+8x^2+8x+x^2-2x-2=0\) Yet again, let's rearrange and see if any combining can take place.
\(x^4-2x^3-4x^3-2x^2+8x^2+x^2+8x-2x-2=0\) Ok, now let's combine.
\(x^4-6x^3+7x^2+6x-2=0\) We can turn this into a function.




Let's check to see that this function fits the strict conditions given in the original problem.


  • The function is monic because the leading coefficient is 1.
  • The function is quartic because the degree is 4.
  • The function is a polynomial because there is no division by a variable and no fractional exponents.
  • The function is written with \(f(x)=\).
  • All coefficients are rational.
  • The function has the given roots 
  • The answer is fully expanded and simplified

This polynomial fits these conditions, so this is the above is the answer.

 Jan 4, 2018

Upon further review, it appears as if this answer is wrong because I copied the wrong roots to start with. The first sentence is already wrong.


"If the roots are \(x=2-\sqrt{3}\)  and \(x=1-\sqrt{3}\), then the conjugates..."


It should be 


"If the roots are \(x=2-\textcolor{red}{3\sqrt{2}}\) and \(x=1-\sqrt{3}\), then the conjugates..."


Unfortunately, I solved the problem using the wrong roots. I encourage you to solve the problem with the example I have given and see if you can do it on your own.

TheXSquaredFactor  Jan 4, 2018

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