How do you multiply $(- 3 + 2 i) (- 3 - 4 i)$ $(-3+2i)(-3-4i)$ ?

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Answer 1 · 2017-09-14T16:34:33

Just like any other polynomials....

...except any time you get a multiple of $i^{2}$ $i^2$ , remove the $i$ $i$ and change the sign on the coefficient.

(-3 + 2i)(-3 - 4i) =

$(- 3 \cdot - 3) + (- 3 \cdot - 4 i) + (2 i \cdot - 3) + (2 i \cdot 4 i)$ $(-3 * -3) + (-3 * -4i) + (2i * -3) + (2i * 4i)$

$= 9 + 12 i - 6 i + 8 i^{2}$ $= 9 + 12i - 6i + 8i^2$

$= 9 + 6 i - 8$ $= 9 + 6i -8$

$= 1 + 6 i$ $= 1 + 6i$

GOOD LUCK!

How do you multiply (−3+2i)(−3−4i)(-3+2i)(-3-4i)?