Answer
a. $(y-2)(y-4)$
b. We cannot factor because there are no pairs of factors of $2$ with a sum of $-1$.
Work Step by Step
a.
$-6$ is negative, so we list negative factors of $8$.
$\left[\begin{array}{lll}
\text{Factors of 8 } & \text{Sum of factors} & \\
-1 \text{ and }-8 & -9 & \\
-2\text{ and } -4 & -6 & \text{..is what we need}
\end{array}\right]$
$y^{2}-6y+8=(y-2)(y-4)$
Check:
$(y-2)(y-4)=$
...use the FOIL method.
$=y^{2}-4y-2y+8$
...add like terms.
$=y^{2}-6y+8$
b.$-1 $ is negative, so we list negative factors of $2$.
$\left[\begin{array}{ll}
\text{Factors of 2 } & \text{Sum of factors} \\
-1\text{ and }-2 & -3
\end{array}\right]$
We cannot factor because there are no pairs of factors of $2$ with a sum of $-1$.
