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Math  /  Algebra

QuestionApply the Distributive Property to show 3i[(2i)+(−3i−5)]3 i[(2 i)+(-3 i-5)] in its simplest form. (1 point)

Studdy Solution

STEP 1

What is this asking? We need to simplify an expression with imaginary numbers using the distributive property and write the result in the simplest form. Watch out! Remember the distributive property and how i2i^2 simplifies!

STEP 2

1. Distribute 3i3i
2. Simplify i2i^2
3. Combine like terms

STEP 3

Let's **distribute** 3i3i across the terms inside the parentheses.
Remember, the distributive property says a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot c.
So, we **multiply** 3i3i by each term inside the square brackets: 3i[(2i)+(−3i−5)]=(3i⋅2i)+(3i⋅(−3i−5))3i[(2i)+(-3i-5)] = (3i \cdot 2i) + (3i \cdot (-3i - 5))

STEP 4

We need to **distribute** 3i3i again to the terms inside the second parentheses: (3i⋅2i)+(3i⋅(−3i−5))=(3i⋅2i)+(3i⋅−3i)+(3i⋅−5)(3i \cdot 2i) + (3i \cdot (-3i - 5)) = (3i \cdot 2i) + (3i \cdot -3i) + (3i \cdot -5)

STEP 5

Now, we **multiply** the terms: (3i⋅2i)+(3i⋅−3i)+(3i⋅−5)=6i2−9i2−15i(3i \cdot 2i) + (3i \cdot -3i) + (3i \cdot -5) = 6i^2 - 9i^2 - 15i

STEP 6

Remember that i=−1i = \sqrt{-1}, so i2=(−1)2=−1i^2 = (\sqrt{-1})^2 = -1.
Let's **substitute** −1-1 for i2i^2: 6i2−9i2−15i=6(−1)−9(−1)−15i6i^2 - 9i^2 - 15i = 6(-1) - 9(-1) - 15i

STEP 7

Now, **multiply** the terms: 6(−1)−9(−1)−15i=−6+9−15i6(-1) - 9(-1) - 15i = -6 + 9 - 15i

STEP 8

Let's **combine** the **constant terms**: −6+9−15i=3−15i-6 + 9 - 15i = 3 - 15i

STEP 9

The simplified form of 3i[(2i)+(−3i−5)]3 i[(2 i)+(-3 i-5)] is 3−15i3 - 15i.

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