How do you factor completely: $2 x^{3} + 14 x^{2} + 6 x + 42$ $2x^3 + 14x^2 + 6x + 42$ ?

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How do you factor completely: $2 x^{3} + 14 x^{2} + 6 x + 42$ $2x^3 + 14x^2 + 6x + 42$ ?

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Answer 1 · 2015-07-17T23:57:40

$2 (x + 7) (x^{2} + 3)$ $2(x+7)(x^2+3)$

Explanation:

Notice that, if you factor out $2 x^{2}$ $2x^2$ from the first two members and factor out $6$ $6$ from the last two members of this equation, you will have the same in both remainders - $(x + 7)$ $(x+7)$ :
$2 x^{3} + 14 x^{2} + 6 x + 42 = 2 x^{2} (x + 7) + 6 (x + 7)$ $2x^3+14x^2+6x+42=2x^2(x+7)+6(x+7)$

Now we can use the distributive law and factor out $(x + 7)$ $(x+7)$ getting:
$(x + 7) (2 x^{2} + 6) = 2 (x + 7) (x^{2} + 3)$ $(x+7)(2x^2+6)=2(x+7)(x^2+3)$

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How do you factor completely: $2 x^{3} + 14 x^{2} + 6 x + 42$ $2x^3 + 14x^2 + 6x + 42$ ?

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Explanation:

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