Solution
= Solution
Expanding the <quadratic form> gives
$$
(x\ y)\begin{pmatrix}1&4\\a_{21}&a_{22}\end{pmatrix}\binom xy
=x^2+(4+a_{21})xy+a_{22}y^2.
$$
Matching <coefficients> of $xy$ and $y^2$ yields
$$
\boxed{a_{21}=4,\qquad a_{22}=4}.
$$
= Solution
Expanding the <quadratic form> gives
$$
(x\ y)\begin{pmatrix}1&4\\a_{21}&a_{22}\end{pmatrix}\binom xy
=x^2+(4+a_{21})xy+a_{22}y^2.
$$
Matching <coefficients> of $xy$ and $y^2$ yields
$$
\boxed{a_{21}=4,\qquad a_{22}=4}.
$$