How do you factor $n^{2} - 10 n + 9$ $n^2-10n+9$ ?

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Answer 1 · 2018-04-10T02:16:25

$(n - 1) (n - 9)$ $(n-1)(n-9)$

$n^{2} - 10 n + 9 = (n - 1) (n - 9)$ $n^2-10n+9 = (n-1)(n-9)$

$n \cdot n = n^{2}$ $n*n=n^2$
$n \cdot - 9 = - 9 n$ $n*-9=-9n$
$n \cdot - 1 = - n$ $n*-1=-n$
$- 9 \cdot - 1 = 9$ $-9*-1=9$

When you combine those you get: $n^{2} - 10 n + 9$ $n^2-10n+9$

How do you factor n2−10n+9n^2-10n+9?