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[tex] \text{Use:}\\(a+b)(c+d)=ac+ad+bc+bd\\i^2=-1\\-------------------------\\(4+7i)(3+4i)=(4)(3)+(4)(4i)+(7i)(3)+(7i)(4i)\\\\=12+16i+21i+28i^2=12+37i+28(-1)=12+37i-28\\=\boxed{-16+37i} [/tex]

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ANSWER

[tex](4 + 7i)(3 + 4i) = - 16 + 37i[/tex]

EXPLANATION

The given expression is
[tex](4 + 7i)(3 + 4i)[/tex]

We need to expand the bracket using the distributive property as follows:

[tex](4 + 7i)(3 + 4i) = 4(3 + 4i) + 7i(3 + 4i)[/tex]

This implies that,

[tex](4 + 7i)(3 + 4i) = 4 \times 3+ 4 \times 4i+ 7i \times 3 + 4i \times 7i[/tex]

We simplify to obtain,

[tex](4 + 7i)(3 + 4i) = 12+ 16i+ 21i + 28 {i}^{2} [/tex]


We now group like terms to obtain,

[tex](4 + 7i)(3 + 4i) = 12+ 37i+ 28 {i}^{2} [/tex]

Now recall that
[tex] {i}^{2} = - 1[/tex]

Applying this property gives,

[tex](4 + 7i)(3 + 4i) = 12+ 28 ( -1 )+ 37i [/tex]

[tex](4 + 7i)(3 + 4i) = 12 - 28+ 37i [/tex]

[tex](4 + 7i)(3 + 4i) = - 16 + 37i[/tex]

The correct answer is A.

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