Math / Algebra

Questioni. Fud the product of this expression. $(3+2 i)(2+6 i)$
2. Simplifut the aven expression $5(2-3 i)$
ACT Questions
3. It you multiply. thes expression. What is the result in follumethod? $\begin{array}{l} (4+5 i)(4-5 i) \\ a \cdot 8+25 i^{2} \\ \text { b. }-8-10 i^{2} \\ c \cdot 16-25 i^{2} \\ d \cdot 16+25 i^{2} \end{array}$

Studdy Solution

STEP 1

What is this asking? We've got to multiply some complex numbers and simplify an expression involving a complex number!
One of the multiplications is a special case we should watch out for. Watch out! Remember that $i^2 = -1$ , not just $i$ !
Also, complex number multiplication distributes just like regular multiplication.

STEP 2

1. Multiply the first expression.
2. Simplify the given expression.
3. Multiply the third expression and find the matching answer.

STEP 3

Let's distribute the terms of the first expression, just like we would with any two binomials: $(3 + 2i)(2 + 6i) = 3 \cdot (2 + 6i) + 2i \cdot (2 + 6i)$ .
This is because multiplication distributes over addition.

STEP 4

Now, let's multiply each term: $3 \cdot 2 = \mathbf{6}$ , $3 \cdot 6i = \mathbf{18i}$ , $2i \cdot 2 = \mathbf{4i}$ , and $2i \cdot 6i = \mathbf{12i^2}$ .

STEP 5

Putting it all together, we have $6 + 18i + 4i + 12i^2$ .
Combining like terms, we get $6 + 22i + 12i^2$ .

STEP 6

Remember that $i^2 = -1$ , so we can replace $12i^2$ with $12(-1) = \mathbf{-12}$ .
Our expression becomes $6 + 22i - 12$ .

STEP 7

Finally, combine the real parts to get $\mathbf{-6 + 22i}$ .
Boom!

STEP 8

We need to simplify $5(2 - 3i)$ .
We distribute the 5 to both terms inside the parentheses.
This gives us $5 \cdot 2 - 5 \cdot 3i$ .

STEP 9

Multiplying gives us $\mathbf{10 - 15i}$ .
That's it for this one!

STEP 10

We have $(4 + 5i)(4 - 5i)$ .
Distributing gives us $4(4 - 5i) + 5i(4 - 5i)$ .

STEP 11

Multiplying each term out gives us $4 \cdot 4 = \mathbf{16}$ , $4 \cdot (-5i) = \mathbf{-20i}$ , $5i \cdot 4 = \mathbf{20i}$ , and $5i \cdot (-5i) = \mathbf{-25i^2}$ .

STEP 12

Putting it all together, we have $16 - 20i + 20i - 25i^2$ .
Notice that $-20i$ and $20i$ add to zero, leaving us with $16 - 25i^2$ .

STEP 13

Since $i^2 = -1$ , we have $16 - 25(-1) = 16 + 25$ .

STEP 14

Adding gives us $\mathbf{41}$ .
This matches answer choice c, $16 - 25i^2$ .

STEP 15

1. $-6 + 22i$
2. $10 - 15i$
3. c

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Studdy Solution

STEP 1

STEP 2

1. Multiply the first expression.
2. Simplify the given expression.
3. Multiply the third expression and find the matching answer.

STEP 3

Let's distribute the terms of the first expression, just like we would with any two binomials: $(3 + 2i)(2 + 6i) = 3 \cdot (2 + 6i) + 2i \cdot (2 + 6i)$ .
This is because multiplication distributes over addition.

STEP 4

Now, let's multiply each term: $3 \cdot 2 = \mathbf{6}$ , $3 \cdot 6i = \mathbf{18i}$ , $2i \cdot 2 = \mathbf{4i}$ , and $2i \cdot 6i = \mathbf{12i^2}$ .

STEP 5

Putting it all together, we have $6 + 18i + 4i + 12i^2$ .
Combining like terms, we get $6 + 22i + 12i^2$ .

STEP 6

Remember that $i^2 = -1$ , so we can replace $12i^2$ with $12(-1) = \mathbf{-12}$ .
Our expression becomes $6 + 22i - 12$ .

STEP 7

Finally, combine the real parts to get $\mathbf{-6 + 22i}$ .
Boom!

STEP 8

We need to simplify $5(2 - 3i)$ .
We distribute the 5 to both terms inside the parentheses.
This gives us $5 \cdot 2 - 5 \cdot 3i$ .

STEP 9

Multiplying gives us $\mathbf{10 - 15i}$ .
That's it for this one!

STEP 10

We have $(4 + 5i)(4 - 5i)$ .
Distributing gives us $4(4 - 5i) + 5i(4 - 5i)$ .

STEP 11

Multiplying each term out gives us $4 \cdot 4 = \mathbf{16}$ , $4 \cdot (-5i) = \mathbf{-20i}$ , $5i \cdot 4 = \mathbf{20i}$ , and $5i \cdot (-5i) = \mathbf{-25i^2}$ .

STEP 12

Putting it all together, we have $16 - 20i + 20i - 25i^2$ .
Notice that $-20i$ and $20i$ add to zero, leaving us with $16 - 25i^2$ .

STEP 13

Since $i^2 = -1$ , we have $16 - 25(-1) = 16 + 25$ .

STEP 14

Adding gives us $\mathbf{41}$ .
This matches answer choice c, $16 - 25i^2$ .

STEP 15

1. $-6 + 22i$
2. $10 - 15i$
3. c

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Studdy Solution

STEP 1

STEP 2

1. Multiply the first expression.2. Simplify the given expression.3. Multiply the third expression and find the matching answer.

STEP 3

Let's **distribute** the terms of the first expression, just like we would with any two binomials: (3+2i)(2+6i)=3⋅(2+6i)+2i⋅(2+6i)(3 + 2i)(2 + 6i) = 3 \cdot (2 + 6i) + 2i \cdot (2 + 6i)(3+2i)(2+6i)=3⋅(2+6i)+2i⋅(2+6i).This is because multiplication distributes over addition.

STEP 4

Now, let's multiply each term: 3⋅2=63 \cdot 2 = \mathbf{6}3⋅2=6, 3⋅6i=18i3 \cdot 6i = \mathbf{18i}3⋅6i=18i, 2i⋅2=4i2i \cdot 2 = \mathbf{4i}2i⋅2=4i, and 2i⋅6i=12i22i \cdot 6i = \mathbf{12i^2}2i⋅6i=12i2.

STEP 5

Putting it all together, we have 6+18i+4i+12i26 + 18i + 4i + 12i^26+18i+4i+12i2.Combining like terms, we get 6+22i+12i26 + 22i + 12i^26+22i+12i2.

STEP 6

Remember that i2=−1i^2 = -1i2=−1, so we can replace 12i212i^212i2 with 12(−1)=−1212(-1) = \mathbf{-12}12(−1)=−12.Our expression becomes 6+22i−126 + 22i - 126+22i−12.

STEP 7

Finally, combine the real parts to get −6+22i\mathbf{-6 + 22i}−6+22i.Boom!

STEP 8

We need to simplify 5(2−3i)5(2 - 3i)5(2−3i).We **distribute** the 5 to both terms inside the parentheses.This gives us 5⋅2−5⋅3i5 \cdot 2 - 5 \cdot 3i5⋅2−5⋅3i.

STEP 9

Multiplying gives us 10−15i\mathbf{10 - 15i}10−15i.That's it for this one!

STEP 10

We have (4+5i)(4−5i)(4 + 5i)(4 - 5i)(4+5i)(4−5i).Distributing gives us 4(4−5i)+5i(4−5i)4(4 - 5i) + 5i(4 - 5i)4(4−5i)+5i(4−5i).

STEP 11

Multiplying each term out gives us 4⋅4=164 \cdot 4 = \mathbf{16}4⋅4=16, 4⋅(−5i)=−20i4 \cdot (-5i) = \mathbf{-20i}4⋅(−5i)=−20i, 5i⋅4=20i5i \cdot 4 = \mathbf{20i}5i⋅4=20i, and 5i⋅(−5i)=−25i25i \cdot (-5i) = \mathbf{-25i^2}5i⋅(−5i)=−25i2.

STEP 12

Putting it all together, we have 16−20i+20i−25i216 - 20i + 20i - 25i^216−20i+20i−25i2.Notice that −20i-20i−20i and 20i20i20i add to zero, leaving us with 16−25i216 - 25i^216−25i2.

STEP 13

Since i2=−1i^2 = -1i2=−1, we have 16−25(−1)=16+2516 - 25(-1) = 16 + 2516−25(−1)=16+25.

STEP 14

Adding gives us 41\mathbf{41}41.This matches answer choice *c*, 16−25i216 - 25i^216−25i2.

STEP 15

1. −6+22i-6 + 22i−6+22i2. 10−15i10 - 15i10−15i3. c

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1. Multiply the first expression.
2. Simplify the given expression.
3. Multiply the third expression and find the matching answer.

Let's distribute the terms of the first expression, just like we would with any two binomials: $(3 + 2i)(2 + 6i) = 3 \cdot (2 + 6i) + 2i \cdot (2 + 6i)$ .
This is because multiplication distributes over addition.

Now, let's multiply each term: $3 \cdot 2 = \mathbf{6}$ , $3 \cdot 6i = \mathbf{18i}$ , $2i \cdot 2 = \mathbf{4i}$ , and $2i \cdot 6i = \mathbf{12i^2}$ .

Putting it all together, we have $6 + 18i + 4i + 12i^2$ .
Combining like terms, we get $6 + 22i + 12i^2$ .

Remember that $i^2 = -1$ , so we can replace $12i^2$ with $12(-1) = \mathbf{-12}$ .
Our expression becomes $6 + 22i - 12$ .

Finally, combine the real parts to get $\mathbf{-6 + 22i}$ .
Boom!

We need to simplify $5(2 - 3i)$ .
We distribute the 5 to both terms inside the parentheses.
This gives us $5 \cdot 2 - 5 \cdot 3i$ .

Multiplying gives us $\mathbf{10 - 15i}$ .
That's it for this one!

We have $(4 + 5i)(4 - 5i)$ .
Distributing gives us $4(4 - 5i) + 5i(4 - 5i)$ .

Multiplying each term out gives us $4 \cdot 4 = \mathbf{16}$ , $4 \cdot (-5i) = \mathbf{-20i}$ , $5i \cdot 4 = \mathbf{20i}$ , and $5i \cdot (-5i) = \mathbf{-25i^2}$ .

Putting it all together, we have $16 - 20i + 20i - 25i^2$ .
Notice that $-20i$ and $20i$ add to zero, leaving us with $16 - 25i^2$ .

Since $i^2 = -1$ , we have $16 - 25(-1) = 16 + 25$ .

Adding gives us $\mathbf{41}$ .
This matches answer choice c, $16 - 25i^2$ .

1. $-6 + 22i$
2. $10 - 15i$
3. c