How do you factor $6 m^{3} - 18 m^{2} - 240 m ?$ $6m ^ { 3} - 18m ^ { 2} - 240m?$

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How do you factor $6 m^{3} - 18 m^{2} - 240 m ?$ $6m ^ { 3} - 18m ^ { 2} - 240m?$

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Answer 1 · 2018-05-02T14:01:25

$6 m (m - 8) (m + 5)$ $6m(m-8)(m+5)$

Explanation:

$take out a common factor 6 m$ $"take out a "color(blue)"common factor "6m$

$= 6 m (m^{2} - 3 m - 40)$ $=6m(m^2-3m-40)$

$the factors of - 40 which sum to - 3 are - 8 and + 5$ $"the factors of - 40 which sum to - 3 are - 8 and + 5"$

$= 6 m (m - 8) (m + 5)$ $=6m(m-8)(m+5)$

Answer 2 · 2018-05-02T14:03:11

$(6 m) (m - 8) (m + 5)$ $(6m)(m-8)(m+5)$

Explanation:

First, notice all terms have $m$ $m$ , so that can be factored out:
$6 m^{3} - 18 m^{2} - 240 m$ $6m^3-18m^2-240m$
$= m (6 m^{2} - 18 m - 240)$ $=m(6m^2-18m-240)$
$6$ $6$ , $18$ $18$ , and $240$ $240$ These numbers' GCF is $6$ $6$ , so we can factor that out aswell:
$= 6 m (m^{2} - 3 m - 40)$ $=6m(m^2-3m-40)$ Now we can factor the parentheses:
$m^{2} - 3 m - 40$ $m^2-3m-40$
Factors of 40:
$1, 40$ $1,40$
$2, 20$ $2,20$
$4, 10$ $4,10$
$5, 8$ $5,8$
$5$ $5$ and $8$ $8$ can be subtracted to get $- 3$ $-3$ , so these are the factors:
$(m - 8) (m + 5)$ $(m-8)(m+5)$ Putting it in the entire expression:
$(6 m) (m - 8) (m + 5)$ $(6m)(m-8)(m+5)$

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How do you factor $6 m^{3} - 18 m^{2} - 240 m ?$ $6m ^ { 3} - 18m ^ { 2} - 240m?$

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How do you factor 6m3−18m2−240m?6m ^ { 3} - 18m ^ { 2} - 240m?

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How do you factor $6 m^{3} - 18 m^{2} - 240 m ?$ $6m ^ { 3} - 18m ^ { 2} - 240m?$