Find the product. Express your answer in standard form.2i(9−4i)(6+5i)2i\left(9-4i\right)\left(6+5i\right)

− 42 + 148 i -42+148i − 42 + 148 i

Find the product. Express your answer in standard form. 2 i ( 9 − 4 i ) ( 6 + 5 i ) 2i\left(9-4i\right)\left(6+5i\right)

Hey, everyone. Let's take a look at this practice problem. So here I wanna find the product and, of course, express my answer in standard form. Now here I actually have 3 complex numbers being multiplied by each other. I have 2 I times 9 minus 4 I times 6 plus 5 I. Now we can still use our steps for multiplication here, we're just gonna start by taking our first two complex numbers and then bring in our third one later. So let's start with step 1, which is to distribute or foil. Since my 2 I is just a single term, I can go ahead and distribute it into this 9 minus 4 I. 2 I times 9 will give me 18 I, and then 2 I times negative 4i will give me negative 8i squared. Now step 2 is to apply I squared equals negative 1. And I have this I squared attached to that negative 8, so this just becomes negative 8 times negative 1, which is just positive 8. Then I can bring my 18i down. I've completed step 2. And step 3 is to combine like terms, but I don't have any like terms that need to get combined here. So now I can go ahead and bring in my 3rd complex number. So all 3 of my steps are done for those first two. Let's restart bringing in our 3rd complex number. So I'm gonna bring this 6+5i down and multiply it by that 18i+8. Now you can go ahead and rewrite this so the first number is in standard form if you wanna see it how you are used to seeing it, but it will still work either way. So restarting with step 1 here, we want to distribute or foil for these numbers. Since they both have two terms, I actually need to go ahead and foil. So let's go ahead and take our first term. 8 times 6 is gonna give me 48. And then my outside terms, 8 times 5 I is gonna give me positive 40 I. Then my inside terms, 18 I times 6 is going to give me positive 108 I. You might have to break out your calculator for these bigger numbers here. And then my last terms, I have 18 I times 5 I, which will give me a positive 90 I squared. So step 1 is done. Moving on to step 2, applying I squared equals negative 1. So I have my I squared term here attached to my 90. So So this becomes 90 times negative one, which is negative 90. So now I can bring all of my other terms down, bringing my 48 and then my 40 I and my 108 I. And from here, we can move on to step 3, which is going to be to combine our like terms. So I first have my 48 and my negative 90, which are gonna combine to give me negative 42. And then I have both of my I terms, 40 I and 108 I, which if I add them together, if I combine them, that gives me positive 148i. Now we want to make sure this is in standard form, of course, a+bi, and it is. So this is my solution, negative 42 +148i. Let me know if you have any questions.

Find the product. Express your answer in standard form. 2 i ( 9 − 4 i ) ( 6 + 5 i ) 2i\left(9-4i\right)\left(6+5i\right)

− 42 + 148 i -42+148i − 42 + 148 i

11. Roots, Radicals, and Complex Numbers

Multiplying and Dividing Complex Numbers

Beginning & Intermediate Algebra

11. Roots, Radicals, and Complex Numbers

Multiplying and Dividing Complex Numbers

Struggling with Beginning & Intermediate Algebra?

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Multiple Choice

Find the product. Express your answer in standard form.
$2 i (9 - 4 i) (6 + 5 i)$

$8+18i$

$54-20i$

$54-40i$

$-42+148i$

Verified step by step guidance

First, recognize that the expression involves multiplying three factors: \$2i$, $(9 - 4i)$, and $(6 + 5i)$. We will multiply them step-by-step.

Start by multiplying the two complex binomials $(9 - 4i)$ and $(6 + 5i)$ using the distributive property (FOIL method): \[ (9 - 4i)(6 + 5i) = 9 \times 6 + 9 \times 5i - 4i \times 6 - 4i \times 5i \]

Simplify each term: \[ 54 + 45i - 24i - 20i^2 \] Recall that $i^2 = -1$, so replace $i^2$ with $-1$ to simplify further.

Combine like terms (real parts and imaginary parts): Real part: \$54 - 20(-1) = 54 + 20$ Imaginary part: \$45i - 24i = 21i$ So the product of the two binomials is: \[ 74 + 21i \]

Now multiply this result by the remaining factor \$2i$: \[ 2i \times (74 + 21i) = 2i \times 74 + 2i \times 21i = 148i + 42i^2 \] Again, replace $i^2$ with $-1$ and simplify the expression to get the final product in standard form (a + bi).

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