How do you factor the monomial $-35a^3c^2$ completely?

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Answer 1 · 2017-03-17T06:45:35

$-1 xx 5 xx 7 xx a xx a xx a xx c xx c$

The negative is a factor. 1 is a factor. 35 can be factored into $5xx7$

$a^3$ = $a xx a xx a$ a times itself three times.

$c^2$ = $c xx c" "$ this is c times itself two times.

How do you factor the monomial -35a^3c^2-35a^3c^2 completely?