# Squaring a Trinomial

What will we get if we expand a general *trinomial*?

\({\left( {x + y + z} \right)^2} = ?\)

We have

\({\left( {x + y + z} \right)^2} = \left( {x + y + z} \right)\left( {x + y + z} \right)\)

Now, we multiply the two brackets term-by-term:

\[\begin{array}{l}{\left( {x + y + z} \right)^2} = \left( {x + y + z} \right)\left( {x + y + z} \right)\\ \qquad\qquad\qquad\! \! = \left\{ \begin{array}{l}x \times \left( {x + y + z} \right)\\ \qquad\qquad+ \\y \times \left( {x + y + z} \right)\\ \qquad\qquad+ \\z \times \left( {x + y + z} \right)\end{array} \right.\\ \qquad\qquad\qquad\! \! = \left\{ \begin{array}{l}{x^2} + xy + xz + \\xy + {y^2} + yz + \\xz + yz + {z^2}\end{array} \right.\end{array}\]

Thus,

\({\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + xz} \right)\)

This is again an identity, an equality which holds true for every value of *x*, *y* and *z*.