How do you factor $n^2-8n+15$ ?

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Answer 1 · 2017-05-19T20:14:45

$(n - 3)(n - 5)$

First of all:

$(n +- x)(n +- y)$

We already have the first term, $n^2$ .

Now, $x$ times $y$ needs to equal $15$ . The factors or $15$ are:

$1, 3, 5, 15$

Since the sum of $3$ and $5$ equal $8$ , which is the coefficient of the middle term, $x = 3$ and $y = 5$ or vice versa.

Since the middle term is negative, both $x$ and $y$ are negative.

The final result is:

$(n - 3)(n - 5)$

How do you factor n^2-8n+15n^2-8n+15?