## Algebra 1: Common Core (15th Edition)

a. $(y-2)(y-4)$ b. We cannot factor because there are no pairs of factors of $2$ with a sum of $-1$.
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 $-1$.