Factor completely $y^2+6y+8$.

Determine the factor.

If $x=a$ is zero of some polynomial $\mathrm{P}\left(x\right)$ then $x-a$ is the Factor of a polynomial such that $x-a$ divides $\mathrm{P}\left(x\right)$ to leave the remainder $\mathrm{zero}$.

Use splitting the middle term method to factor the given polynomial:

$\begin{array}{rcl}{y}^2+6y+8& =& y^2+(4+2)y+8\\ & =& y^2+4y+2y+8\\ & =& y(y+4)+2(y+4)\\ & =& (y+4)(y+2)\end{array}$

Hence, the factor of ${\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{+}\mathbf{8}$ is $\mathbf{(}\mathit{y}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{}\mathrm{and}\mathbf{(}\mathit{y}\mathbf{+}\mathbf{2}\mathbf{)}$.