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Question

$(i x - 5) (6 + 2 i x) (5 - 2 i) = 0$ $(ix-5)(6+2ix)(5-2i)=0$

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Answer 1 · 2018-01-10T19:19:45

A product is equal to $0$ $0$ if and only if at least one of the factors is $0$ $0$ . The left side is already factored for us.

Solve $(i x - 5) = 0$ $(ix-5)=0$ or $(6 + 2 i x) = 0$ $(6+2ix) = 0$ (It is clear that $(5 - 2 i)$ $(5-2i)$ cannot be $0$ $0$ )

$x = \frac{5}{i} = - 5 i$ $x = 5/i = -5i$ or $x = \frac{- 6}{2 i} = \frac{- 3}{i} = 3 i$ $x = (-6)/(2i) = (-3)/i = 3i$

Check the answers by substitution.