Question

If $p\left(y\right)=y^2-3\sqrt{2}y+1$ then find $p\left(3\sqrt{2}\right)$.

Answer 1

$p\left(y\right)=y^2-3\sqrt{2}y+1$

$\therefore p\left(3\sqrt{2}\right)={\left(3\sqrt{2}\right)}^2-3\sqrt{2}\times 3\sqrt{2}+1=18-18+1=1$