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

2 Answers
Aug 8, 2015

Answer:

#(y-4)(y-8)#

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 and 32 -> 33# [X]
#2 and 16 -> 18# [X]
#4 and 8 -> 12# [#sqrt#]

so it looks like the factor pair of #4 and 8# works!

we would just substitute the two numbers in for the two "#?#"
and get
#(y-4)(y-8)#

Aug 8, 2015

Answer:

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)