How do you factor $3 n^{2} - 8 n + 5$ $3n^2-8n+5$ ?

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Answer 1 · 2018-03-31T16:13:49

$(n - 1) (3 n - 5)$ $(n-1)(3n-5)$

1.) write $- 8 n$ $-8n$ as a sum or difference
$\Rightarrow 3 n^{2} - 3 n - 5 n + 5$ $=>3n^2-3n-5n +5$

2.) factor out $3 n$ $3n$ and $- 5$ $-5$ from the expression:
$\Rightarrow 3 n (n - 1) - 5 (n - 1)$ $=>3n (n-1)-5(n-1)$

3.) factor out $(n - 1)$ $(n-1)$ from the expression
$\Rightarrow (n - 1) (3 n - 5)$ $=>(n-1)(3n-5)$

Answer 2 · 2018-03-31T16:23:24

$(3 n - 5) (n - 1)$ $(3n - 5)(n - 1)$

$3 n^{2}$ $3n^2$ only has two factors, $3 n and n$ $3n and n$

5 only has two factors , 5 and 1. These are both negative to multiply to +5 and add to make -8

How do you factor 3n2−8n+53n^2-8n+5?