Iwant to do this problem in product standard form (5i-3)(2i+1)

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Work Shown:

Let x = 5i-3

That allows us to go from (5i-3)(2i+1) to x(2i+1)

Distribute the x through: x*(2i) + x*(1) = 2i*x + x = 2i(x) + 1(x)

Now we replace every x in 2i(x) + 1(x) with 5i-3, and then we distribute a second time

2i(x) + 1(x) = 2i(5i-3) + 1(5i-3)

2i(x) + 1(x) = 2i(5i)+2i(-3) + 1(5i)+1(-3)

2i(x) + 1(x) = 10i^2 - 6i + 5i - 3

2i(x) + 1(x) = 10(-1) - i - 3

2i(x) + 1(x) = -10 - i - 3

2i(x) + 1(x) = -13 - i

Therefore, (5i-3)(2i+1) = -13 - i

The result is in a+bi form where a = -13 and b = -1.

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An alternative method is to use the box method. This is where you set up a grid that helps you multiply out (5i-3)(2i+1)

See the diagram below.

Each of the 4 red terms in the boxes represents the result of multiplying the outer blue and green terms. Example: 5i times 2i = 10i^2 in row1, column1.

i got 10.28 but that rounds up to 10.3