How do you factor $3 p^{2} - 2 p - 5$ $3p^2 - 2p - 5$ ?

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Answer 1 · 2015-06-20T20:07:09

$3 p^{2} - 2 p - 5 = (p + 1) (3 p - 5)$ $3p^2-2p-5 = (p+1)(3p-5)$

Let $f (p) = 3 p^{2} - 2 p - 5$ $f(p) = 3p^2-2p-5$ .

Notice that $f (- 1) = 3 + 2 - 5 = 0$ $f(-1) = 3+2-5 = 0$ .

So $(p + 1)$ $(p+1)$ is a factor of $f (p)$ $f(p)$

In order to get the $3 p^{2}$ $3p^2$ and the $- 5$ $-5$ terms, the other factor must be $(3 p - 5)$ $(3p-5)$ so

$3 p^{2} - 2 p - 5 = (p + 1) (3 p - 5)$ $3p^2-2p-5 = (p+1)(3p-5)$

How do you factor 3p2−2p−53p^2 - 2p - 5?