# How do you factor completely: y² -  12y  +  32?

Aug 8, 2015

$\left(y - 4\right) \left(y - 8\right)$

#### Explanation:

because the middle sign is a minus
and the last sign is an addition, both the signs in the parenthesis would be minuses
(y-?)(y-?)

now the two "?" numbers
will be a pair of factors of $32$
and add up to $12$

so let's list the pairs of factors of $32$ and what they add up to :)
$1 \mathmr{and} 32 \to 33$ [X]
$2 \mathmr{and} 16 \to 18$ [X]
$4 \mathmr{and} 8 \to 12$ [sqrt]

so it looks like the factor pair of $4 \mathmr{and} 8$ works!

we would just substitute the two numbers in for the two "?"
and get
$\left(y - 4\right) \left(y - 8\right)$

Aug 8, 2015

Solve f(y) = y^2 - 12y + 32

Ans: (y - 4)(y - 8)

#### Explanation:

Find 2 numbers p and q knowing sum (-12) and product (32).
p and q have same sign, since a and c same sign.
Factor pairs of 32 --> (2, 16)(4, 8). This sum is 12 = - b.
Then, p = -4 and q = -8 (their sum must be -12, and not 12)

f(y) = (y - 4)(y - 8)