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

Answer 1

See below:

#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]}

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Answer 2

Hazel Anglin

To factor the expression y = n^2 - 16n + 64, it factors into the perfect square trinomial (n - 8)^2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.