MATH SOLVE

4 months ago

Q:
# Multiply? I don't know how to do this o.o(X+4β2) (x-4β2) (x-3i)

Accepted Solution

A:

Answer:[tex]x^{3} -(3i) \; x^{2} -32x + 96 i[/tex].Step-by-step explanation:Notice that the first two factors are in the form [tex](x-a)(x+a)[/tex], which is equal to [tex](x^{2} - a^{2})[/tex]. Start by combining and expanding these two factors:Let [tex]a = 4\sqrt{2}[/tex].[tex]a^{2} = 16 \times 2 = 32[/tex].[tex](x + a) (x - a) = x^{2} - a^{2} = x^{2} -32[/tex].This expression can now be expressed as [tex](x^{2} - 32)(x - 3i)[/tex]. [tex]i[/tex] stands the unit imaginary number, where [tex]i^{2} = -1[/tex]. Unless [tex]i[/tex] is raised to a certain power other than [tex]1[/tex], it can be treated just like a constant.Expand this expression using FOIL:[tex]\begin{aligned}&(x^{2} - 32)(x - 3i)\\=&\underbrace{x^{2}\cdot x}_{\verb!F!} +\underbrace{(x^{2})(-3i)}_{\verb!O!} + \underbrace{(-32)x}_{\verb!I!} + \underbrace{(-32)(-3i)}_{\verb!L!}\\=& x^{3} -(3i)x^{2}-32x + 96i \end{aligned}[/tex].