In Exercises 65–70, perform the indicated operation(s) and write ...

Question

In Exercises 65–70, perform the indicated operation(s) and write the result in standard form.(8 + 9i)(2 - i) - (1 - i)(1 + i)

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we want to simplify the expression six plus five I times three minus I minus three minus I times three plus I. And we want to write the answer in standard form. So in this case you can see that there's two sort of components to this expression. We have this first set of parentheses and then we have this second set of parentheses. So if I look at just the first set of parentheses, I have six plus five I times three minus I And because they're parentheses right next to each other I need to multiply them with one another which means I want to multiply each component in the first parentheses to each component in the second parentheses. So I can use the foil method which means the first terms get multiplied. So six times three is 18 than the outer terms six times negative I is negative six I and then the inner terms five I times three is positive 15 I and five I times negative I is negative five I squared. And I can combine these middle terms leaving me with plus nine I minus five I squared. Well I know that I is equal to the square root of negative one. So if I have I squared I have the square root of negative one times the square root of negative one. Which means I have a square root of -1 squared. So I squared can be simplified or expressed as -1. So if I go back to this expression where I have negative five I squared, I can write this as 18 plus nine, I minus five times negative one. Where I replaced the I squared with negative one. If I do this simplification I have 18 plus nine. I negative five times negative one is positive five. So now I can combine the 18 and the five Leaving me with 23 plus nine. I so now I simplified the left hand side of my expression And now I can simplify the right hand side so the right hand side is 3 - times three plus I. And if you notice these are square differences of one another. So If I multiply these together I get three squared plus three I minus three I minus I squared. So those middle terms end up canceling out and I'm left with three squared minus I squared and we've already determined that I squared is equal to negative one. So I can rewrite this as three squared minus negative one. So if I simplify this I have three squared, two negatives make a positive plus one and nine squared is actually nine. So nine plus one is 10. So now I've simplified the left hand side and the right hand side. So I'm gonna go ahead and put them back together. So I have 23 plus nine, I minus the right hand side, 10. So because 10 is a real component, I can subtract it from the 23. so this becomes 23 plus nine I minus 10, and I can subtract 10 from 23 leaving me with 13 plus nine I. So this becomes my final answer when I'm simplifying this expression and you can see that it's already written in standard form because I have the real component first followed by the imaginary and this answer aligns with answer choice A. So hopefully this video helped and see you next time.

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In Exercises 65–70, perform the indicated operation(s) and write the result in standard form.(8 + 9i)(2 - i) - (1 - i)(1 + i)

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