To factor #x^2-17x+72#, note first that because the coefficient (#-17#) of the term in #x# is negative and the constant term #72# is positive that the factorisation must take the form:
#x^2-17x+72 = (x-a)(x-b)#
where #a# and #b# are positive, #a+b = 17# and #ab = 72#
So we want to find #a, b > 0# such that #a+b = 17# and #ab = 72#.
The finding of such #a# and #b# is what makes these problems fun or challenging, depending on your perspective.
Method 1 - Linear Search
Note first that if #a < b# then #a <= sqrt(72)# so we find #a <= 8# since #8^2 = 64 < 72 < 81 = 9^2#.
That means #a = 1, 2, 3, 4, cancel(5), 6, cancel(7) or 8#.
I have crossed out #5# and #7# since they are not factors of #72#.
If we let #b = 72/a# for each of these values, we get the possibilities:
#b = 72, 36, 24, 18, 12 or 9# respectively,
with sums:
#a+b = 73, 38, 27, 22, 18 or 17#
So the last of these is the one we want #a = 8# and #b = 9#.
Method 2 - Prime Factorisation and some reasoning
Note that #72 = 2*2*2*3*3# as a product of primes.
We can find this by noting that #72# is even, so divide by #2# to get #72 = 2*36#. Then #36# is even, so divide that by #2# to get #72=2*2*18#, etc. When we get to #72 = 2*2*2*9#, #9# is no longer even, but easily factorisable into #3*3#.
Having split #72# into these factors, note that finding pairs #(a, b)# such that #ab = 72# can be done by splitting the prime factors into two boxes and multiplying them up. For example, #{2, 2, 3}# and #{2, 3}# giving #72 = 12xx6#.
Next note that #17# is not even, hence all the #2#'s must be in one of the boxes. Otherwise #a#, #b# and #a+b# would all be even.
Similarly, note that #17# is not divisible by #3#. So all of the #3#'s must be in one of the boxes.
That only gives two possible splits, corresponding to #1 xx 72# or #8 xx 9#. The first split gives a sum of #73#, the second the desired value #17#.
