+647 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.
+2446 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.
\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}=0\)
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. |
| \(\textcolor{blue}{x^2-2x-2}\) | |
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. |
\(f(x)=x^4-6x^3+7x^2+6x-2\)
Let's check to see that this function fits the strict conditions given in the original problem.
This polynomial fits these conditions, so this is the above is the answer.
+2446 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.
