Is it possible to factor #y= 16x^2 -24x + 9 #? If so, what are the factors?

1 Answer
Prof_S
Jan 11, 2016

#(4x-3)(4x-3)#

Explanation:

It is a good idea to try integer coefficients in the factors, since the function itself has all integer coefficients. Since #9 = 3^2# and the middle term is negative, it's good to check out factors of the form (ax-3)(bx-3), for some coefficients a and b. Since #ax# times #bx# must give #16x^2#, and #16=2^4#, the coefficients a and b must be powers of 2. Trying possibilities, we see it works for a=4 and b=4. That is, for this choice of powers of 2, we get the middle term, namely -24x.

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