Find a monic quartic polynomial f(x) with rational coefficients whose roots include $x=2-3\sqrt{2}$ and $x=1-\sqrt{3}$.
How could you find the other roots?
[Math] Find monic quartic polynomial f(x) with rational coefficients whose roots include…(Algebra)
polynomialsroots
Best Answer
Assuming the polynomial has real coefficients. Sort of implied. Rational coefficients are real, unless you want to extend the definition to gaussian rationals.
In that case, because all coefficients are rational, all irrational roots of the form $a+\sqrt{b}$, for rational $a,b$ and irrational $\sqrt{b}$ come in conjugate pairs. We can conclude straight away that there are also roots $x=2+3\sqrt{2}$ and $x=1+\sqrt{3}$.
This should be easy enough for you to figure out from there.
Note: It would be incorrect to say that every irrational root has a conjugate pair. Only irrational roots of the form $a+\sqrt{b}$ have a conjugate pair. Its a subtlety, but its a common misconception Ive seen far too often. A quartic could conceivably have one linear factor and one cubic factor, and cubics typically have irrational roots of the form $a+\sqrt[3]{b+\sqrt{c}}+\sqrt[3]{b-\sqrt{c}}$. The rule would not apply.
A helpful tidbit: If you have irrational conjugate roots $a\pm\sqrt{b}$ then you have a single quadratic factor of the form $x^2 -2ax +(a^2 - b)$.