How do you solve $(3 n - 2) (4 n + 1) = 0$ $(3n - 2) (4n + 1) = 0$ by factoring?

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How do you solve $(3 n - 2) (4 n + 1) = 0$ $(3n - 2) (4n + 1) = 0$ by factoring?

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Answer 1 · 2018-03-20T10:33:33

$n = - \frac{1}{4} or n = \frac{2}{3}$ $n=-1/4" or "n=2/3$

Explanation:

$since the product of the factors is zero then one or both of$ $"since the product of the factors is zero then one or both of"$
$the factors must equal zero$ $"the factors must equal zero"$

$to solve equate each factor to zero and solve for n$ $"to solve equate each factor to zero and solve for n"$

$3 n - 2 = 0 \Rightarrow n = \frac{2}{3}$ $3n-2=0rArrn=2/3$

$4 n + 1 = 0 \Rightarrow n = - \frac{1}{4}$ $4n+1=0rArrn=-1/4$

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How do you solve $(3 n - 2) (4 n + 1) = 0$ $(3n - 2) (4n + 1) = 0$ by factoring?

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How do you solve (3n−2)(4n+1)=0(3n - 2) (4n + 1) = 0 by factoring?

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How do you solve $(3 n - 2) (4 n + 1) = 0$ $(3n - 2) (4n + 1) = 0$ by factoring?