How do you factor #y=n^2-16n+64 #?

1 Answer
Apr 11, 2018

Answer:

See below:

Explanation:

#y=n^2-16n+64#

I think the easiest way to think about a problem when asked to factorize is: "What two numbers, when added gives -16, and when multiplied gives 64?"

When factoring in this case you would get:

#(n+x)(n+y)#

But we know that #x+y=-16# and #x times y =64#
And then we can conclude that the number in question must be #-8#.

So the factorized version would be:
#(n-8)(n-8)#

So the quadratic has a repeated solution: #8#

#x=8# is therefore a solution- which can be seen in the graph of the function:
graph{x^2-16x+64 [-10, 10, -5, 5]}