Question

How do you find the polynomial function with roots $2,\;3 \pm \sqrt 2 ?$

Answer 1

Hint: Since all the roots are given for the required polynomial function, so first find the factors of the polynomial function and then find the product of all the factors in order to find the polynomial function. If $ \pm a$ is the root of a polynomial function $f(x)$ then $(x \mp a)$ will be the factor of the given polynomial function.

Complete step by step answer:
In order to find the polynomial function whose all roots are given, we will find first find the factors of the function using the given roots as follows
The given roots of the required polynomial function are $2,\;3 \pm \sqrt 2 $
Here three roots are given, that means we will get three factors,
Factor including the root $2$ will be given as $(x - 2)$
Factor including the root $3 + \sqrt 2 $ will be given as $\left( {x - \left( {3 + \sqrt 2 } \right)} \right)$
Factor including the root $3 - \sqrt 2 $ will be given as $\left( {x - \left( {3 - \sqrt 2 } \right)} \right)$
Now, we will multiply all the factors to get the required polynomial function
So in this way the polynomial function $f(x)$ will be given as
$
  f(x) = (x - 2)\left( {x - \left( {3 + \sqrt 2 } \right)} \right)\left( {x - \left( {3 - \sqrt 2 } \right)} \right) \\
  f(x) = (x - 2)\left( {x - 3 - \sqrt 2 } \right)\left( {x - 3 + \sqrt 2 } \right) \\
  f(x) = (x - 2)\left( {(x - 3) - \sqrt 2 } \right)\left( {(x - 3) + \sqrt 2 } \right) \\
 $
Using the algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$ to simplify further,
\[
  f(x) = (x - 2)\left( {{{(x - 3)}^2} - {{\left( {\sqrt 2 } \right)}^2}} \right) \\
  f(x) = (x - 2)\left( {{x^2} + 9 - 6x - 2} \right) \\
  f(x) = (x - 2)\left( {{x^2} + 7 - 6x} \right) \\
  f(x) = x\left( {{x^2} + 7 - 6x} \right) - 2\left( {{x^2} + 7 - 6x} \right) \\
  f(x) = {x^3} + 7x - 6{x^2} - 2{x^2} - 14 + 12x \\
  f(x) = {x^3} - 8{x^2} +19x - 14 \\
 \]
Therefore \[f(x) = {x^3} - 8{x^2} - 5x + 14\] is the required polynomial function.

Note: This method is not useful in the case where there are identical roots, because you never know how much of there are identical roots in the polynomial if it is not given in the question. Also the number of roots also decides the degree of the polynomial function, as you can see in this question, there are a total of three roots and also the degree of the polynomial function is three.