## Precalculus (6th Edition)

$\dfrac{3}{10}-\dfrac{8}{5}i$
To simplify the given expression, their is a need to rationalize the denominator. This can be done by multiplying the conjugate of the denominator, which is $2-4i$, to both the numerator and the denominator to obtain: $\dfrac{7-2i}{2+4i} \\=\dfrac{(7-2i)(2-4i)}{(2+4i)(2-4i)}$ RECALL: (1) $(a-b)(a+b) = a^2-b^2$ (2) $(a+b)(c+d)=ac +ad + bc+bd$ (popularly known as the FOIL method) Use the rule in (1) above to simplify the denominator and the FOIL method to expand the numerator to obtain: $=\dfrac{7(2)+7(-4i)+(-2i)(2)+(-2i)(-4i)}{2^2-(4i)^2} \\=\dfrac{14-28i-4i+8i^2}{4-16i^2} \\=\dfrac{14-32i+8i^2}{4-16i^2}$ Since $i^2=1$, the expression above simplifies to: $=\dfrac{14-32i+8(-1)}{4-16(-1)} \\=\dfrac{14-32i-8}{4+16} \\=\dfrac{6-32i}{20} \\=\dfrac{2(3-16i)}{2(10)}$ Simplify by canceling the common factor $2$ to obtain: $=\dfrac{3-16i}{10} \\=\dfrac{3}{10} - \dfrac{16}{10}i \\=\dfrac{3}{10}-\dfrac{8}{5}i$