Question

If f(x) is a polynomial of degree n with rational coefficients, and $1+2i,2-\sqrt{3},5$ are three roots of f(x) = 0, then the least value of n is
[a] 5
[b] 4
[c] 0.5
[d] 6

Answer 1

Hint: Use fundamental theorem of algebra, i.e. a polynomial of degree n has exactly n roots over the field of complex numbers. Use the fact that if all the coefficients of a polynomial are rational numbers, then for every irrational root of form $a-\sqrt{b}$ or $a+\sqrt{b},a,b\in \mathbb{Q}$, the radical conjugate of the root $a+\sqrt{b}$ or $a-\sqrt{b}$ respectively is also a root. Also, for every complex root of the form $a+ib$ or $a-ib,a,b\in \mathbb{Q}$, the complex conjugate is also a root. Hence find the minimum number of necessary roots that the polynomial must have in order to have all the coefficients to be rational. Hence using the fundamental theorem of algebra find the minimum degree of the polynomial.

Complete step-by-step answer:
Fundamental theorem of algebra: According to the fundamental theorem of algebra, every polynomial of degree n has no more than n roots and has exactly n complex roots, i.e. over the field of complex numbers a polynomial of degree n has exactly n roots except if the polynomial is a zero polynomial.
Also, if a polynomial has only rational coefficients, then the radical conjugate of every root is also a root of the polynomial and the complex conjugate of every root is also a root of the polynomial.
Now since f(x) has a root $1+2i$, we have f(x) also has the root as $1-2i$
Since f(x) has $2-\sqrt{3}$ as its root, $2+\sqrt{3}$ is also a root of f(x).
Hence the roots of f(x) are $1+2i,1-2i,2+\sqrt{3},2-\sqrt{3},5$
Since there are a minimum of five roots of f(x) the degree of f(x) is at least five according to the fundamental theorem of algebra.
Hence the minimum value of n is 5.
Hence option [a] is correct.

Note: We can also note that the degree of the above polynomial is at least five by making use of factor theorem.
Since $f\left( 1+2i \right)=f\left( 1-2i \right)=f\left( 2+\sqrt{3} \right)=f\left( 2-\sqrt{3} \right)=f\left( 5 \right)=0$, we have
$g\left( x \right)=\left( x-1-2i \right)\left( x-1+2i \right)\left( x-2-\sqrt{3} \right)\left( x-2+\sqrt{3} \right)\left( x-5 \right)$ is a factor of f(x).
We know that if $g\left( x \right)$ is a factor of $f\left( x \right)$, then $\deg \left( f\left( x \right) \right)\ge \deg \left( g\left( x \right) \right)$
Hence we have $\deg \left( f\left( x \right) \right)\ge 5$
Hence we have $n\ge 5$
Hence the minimum value of n is 5.