How do you factor $12 m^{2} n^{2} p^{2} - m n p - 6$ $12m^2n^2p^2-mnp-6$ ?

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How do you factor $12 m^{2} n^{2} p^{2} - m n p - 6$ $12m^2n^2p^2-mnp-6$ ?

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Answer 1 · 2016-05-28T13:55:55

$3 \times 2^{2} \times (m n p + \frac{2}{3}) \times (m n p - \frac{3}{4})$ $3 times 2^2 times (m n p +2/3)times (m n p - 3/4)$

Explanation:

Making $x = m n p$ $x = m n p$ the resulting equation $12 x^{2} - x - 6 = 0$ $12 x^2-x-6=0$
has roots: $x = - \frac{2}{3}$ $x = -2/3$ and $x = \frac{3}{4}$ $x= 3/4$
So we can represent $12 x^{2} - x - 6 = 12 (x + \frac{2}{3}) (x - \frac{3}{4})$ $12 x^2-x-6 = 12(x+2/3)(x-3/4)$ The final factorization is
$3 \times 2^{2} \times (m n p + \frac{2}{3}) \times (m n p - \frac{3}{4})$ $3 times 2^2 times (m n p +2/3)times (m n p - 3/4)$

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How do you factor $12 m^{2} n^{2} p^{2} - m n p - 6$ $12m^2n^2p^2-mnp-6$ ?

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How do you factor $12 m^{2} n^{2} p^{2} - m n p - 6$ $12m^2n^2p^2-mnp-6$ ?