Exercise 26 Page 411

Here we have a quadratic trinomial of the form $a{x}^2+bx+c,$ where there are no common factors. We can see that $a=\text{-}5,$ $b=\text{-}22,$ and $c=\text{-}8.$ Since $c$ is negative, the factors of $c$ must have different signs. Factors of $\text{-}5$ are $1$ and $\text{-}5.$ Example factors of $\text{-}8$ are $1$ and $\text{-}8.$ Let's consider the possible factorization. To see whether the factorization is correct, we multiply and check if the middle term equals the middle term of the given polynomial, $\text{-}22y.$ Now, let's compare middle terms. Let's do the same with the rest of the factors.

Factors of $\text{-}5$	Factors of $\text{-}8$	Possible Factorization	Middle Term
$1,$ $\text{-}5$	$1,$ $\text{-}8$	$(y+1)(\text{-}5y-8)$	$\text{-}8y-5y=\text{-}13y$ $\times$
$1,$ $\text{-}5$	$\text{-}8,$ $1$	$(y-8)(\text{-}5y+1)$	$y+40y=41y$ $\times$
$1,$ $\text{-}5$	$\text{-}1,$ $8$	$(y-1)(\text{-}5y+8)$	$8y+5y=13y$ $\times$
$1,$ $\text{-}5$	$8,$ $\text{-}1$	$(y+8)(\text{-}5y-1)$	$\text{-}y-40y=\text{-}41y$ $\times$
$1,$ $\text{-}5$	$2,$ $\text{-}4$	$(y+2)(\text{-}5y-4)$	$\text{-}4y-10y=\text{-}14y$ $\times$
$1,$ $\text{-}5$	$\text{-}4,$ $2$	$(y-4)(\text{-}5y+2)$	$2y+20y=22y$ $\times$
$1,$ $\text{-}5$	$\text{-}2,$ $4$	$(y-2)(\text{-}5y+4)$	$4y+10y=14y$ $\times$
$1,$ $\text{-}5$	$4,$ $\text{-}2$	$(y+4)(\text{-}5y-2)$	$\text{-}2y-20y=\text{-}22y$ $✓$

Looking at the table, we can finally factor the given polynomial.