If $p(x)=x^{2}-2 \sqrt{2} x+1$, then what is the value of  $p(2 \sqrt{2})$

Given :

The given expression is $p(x)=x^{2}-2 \sqrt{2} x+1$.

To find :

We have to find the value of  $p(2 \sqrt{2})$.

Solution :

$p(x)=x^{2}-2 \sqrt{2} x+1$

Substitute $x = 2\sqrt{2}$

$p(2 \sqrt{2}) = (2 \sqrt{2})^{2}-2 \sqrt{2}(2 \sqrt{2})+1$

                          $ = 4 \times 2 - 4 \times 2 + 1$

                          $ = 8 - 8 + 1 = 0 + 1$

                           $ = 1.$

The value of $p(x)=x^{2}-2 \sqrt{2} x+1$ at $p(2 \sqrt{2})$ is 1.