How do you factor completely $3 w^{2} - 75$ $3w^2-75$ ?

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Answer 1 · 2016-06-03T14:59:10

$3 (w + 5) (w - 5)$ $3(w+5)(w-5)$

To factor $3 w^{2} - 75$ $3w^2-75$

Begin by factoring out the common factor $3$ $3$

$3 (w^{2} - 25)$ $3(w^2-25)$

Now we recognize that $w^{2} - 25$ $w^2-25$ is the difference of two squares

$3 (w + 5) (w - 5)$ $3(w+5)(w-5)$

How do you factor completely 3w2−753w^2-75?