Question

How do you find a fourth degree polynomial given roots \[\sqrt 2 \] and \[2 - \sqrt 3 \]?

Answer 1

Hint: To find a fourth degree polynomial, we will first obtain all four roots and for that we will use the fact that if there is a polynomial with all rational coefficients, then if it has one irrational root, its conjugate will also be a root. So, the roots will be \[\sqrt 2 \], \[ - \sqrt 2 \], \[2 - \sqrt 3 \] and \[\left( {2 + \sqrt 3 } \right)\]. After this we will make two different quadratic equations by using \[ \pm \sqrt 2 \] and \[\left( {2 \pm \sqrt 3 } \right)\]. Finally, we will multiply the two obtained quadratic equations to obtain a fourth degree polynomial.

Complete step by step answer:
We have to find a fourth degree polynomial given roots \[\sqrt 2 \] and \[2 - \sqrt 3 \].Let the fourth degree polynomial be \[p(x)\]. As we know that if a polynomial with all rational coefficients, then if it has one irrational root, its conjugate will also be a root. So, if \[\sqrt 2 \] is a root then \[ - \sqrt 2 \] is also a root. Similarly, if \[2 - \sqrt 3 \] is a root then \[\left( {2 + \sqrt 3 } \right)\] is also a root. Therefore, the roots will be \[\sqrt 2 \], \[ - \sqrt 2 \], \[2 - \sqrt 3 \] and \[\left( {2 + \sqrt 3 } \right)\].

As we know, a quadratic equation is of form:
\[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]
Consider, first two roots \[\sqrt 2 \] and \[ - \sqrt 2 \], we have
\[ \Rightarrow {x^2} - \left( {\sqrt 2 - \sqrt 2 } \right)x + \left( {\left( {\sqrt 2 } \right)\left( { - \sqrt 2 } \right)} \right) = 0\]
On simplifying, we get
\[ \Rightarrow {x^2} - 2 = 0 - - - (1)\]
Now, consider the other two roots \[2 - \sqrt 3 \] and \[\left( {2 + \sqrt 3 } \right)\].
\[ \Rightarrow {x^2} - \left( {2 - \sqrt 3 + 2 + \sqrt 3 } \right)x + \left( {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right) = 0\]

On simplifying and using the identity: \[{a^2} - {b^2} = (a - b)(a + b)\], we get
\[ \Rightarrow {x^2} - 4x + 1 = 0 - - - (2)\]
Multiplying \[(1)\] and \[(2)\], we get \[p(x)\].
 \[ \Rightarrow p(x) = \left( {{x^2} - 2} \right)\left( {{x^2} - 4x + 1} \right)\]
On expanding, we get
\[ \Rightarrow p(x) = {x^4} - 4{x^3} + {x^2} - 2{x^2} + 8x - 2\]
On simplification, we get
\[ \therefore p(x) = {x^4} - 4{x^3} - {x^2} + 8x - 2\]

Therefore, a fourth degree polynomial given roots \[\sqrt 2 \] and \[2 - \sqrt 3 \] is \[{x^4} - 4{x^3} - {x^2} + 8x - 2\].

Note: If a polynomial with all rational coefficients, then if it has one irrational root, its conjugate will also be a root of the given polynomial. Similarly, imaginary roots or complex roots always occur in pairs. For example, if given \[a + ib\] is a root then \[a - ib\] will also be the root of the polynomial.