Find each product. Write answers in standard form. 3i(2-i)²

Question

Accepted Answer

Hello, everyone. We're going to do multiplication to be able to write five I times seven minus I squared in standard form. The first thing I wanna do is do out the seven minus I squared. If you recall when you have a binomial squared, you square. The first term. So seven squared gives me 49 square. The last term negative I squared gives me plus I squared and take the product of the two terms and double it. So the product of seven times negative I is negative seven I doubled is negative 14 I. So now that that's out of the way there's a couple of things I could do next, what I'm gonna do is I want to take care of that neg of that I squared I squared is the same as negative one. So that's what I'm gonna do. I'm gonna replace I squared with negative one. So this is now plus a negative one and I can combine my like terms within my parentheses. So the 49 and the negative one can be combined. So I now have five I times in parentheses, 48 minus 14. I now this will be a little simpler in the next step. So we'll distribute the five I into the parentheses. Five times 48 gives me 240 the I ends up there five I times negative 14, I will give me a negative 70 I squared and just like I did in the previous part, I'm gonna replace I squared with negative one. So 240 I minus 70 times negative one. So minus 70 times a negative one will give me a plus 70. So I now have 240 I plus 70. But in the interest of writing this in standard form, in standard form, we put the real part first. So the real part is the 70 plus the imaginary part which is 240 I. So our solution is 70 plus 240 I which is answer choice B have a nice day.

Find each product. Write answers in standard form. 3i(2-i)²

Key Concepts

Complex Numbers

Exponentiation

Standard Form of Complex Numbers

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