15 points and brainliest
what is 1-49m^2 factored?

Step-by-step explanation:

A complex number is any number that can be written as \greenD{a}+\blueD{b}ia+bistart color #1fab54, a, end color #1fab54, plus, start color #11accd, b, end color #11accd, i, where iii is the imaginary unit and \greenD{a}astart color #1fab54, a, end color #1fab54 and \blueD{b}bstart color #11accd, b, end color #11accd are real numbers.

When multiplying complex numbers, it's useful to remember that the properties we use when performing arithmetic with real numbers work similarly for complex numbers.

Sometimes, thinking of iii as a variable, like xxx, is helpful. Then, with just a few adjustments at the end, we can multiply just as we'd expect. Let's take a closer look at this by walking through several examples.

Multiplying a real number by a complex number

Example

Multiply -4 (13+5i)−4(13+5i)minus, 4, left parenthesis, 13, plus, 5, i, right parenthesis. Write the resulting number in the form of a+bia+bia, plus, b, i.

Solution

If your instinct tells you to distribute the -4−4minus, 4, your instinct would be right! Let's do that!

\begin{aligned}\tealD{-4}(13+5i)&=\tealD{-4}(13)+\tealD{(-4)}(5i)\\ \\ &=-52-20i \end{aligned}

−4(13+5i)

​

=−4(13)+(−4)(5i)

=−52−20i

​

And that's it! We used the distributive property to multiply a real number by a complex number. Let's try something a little more complicated.

Multiplying a pure imaginary number by a complex number

Example

Multiply 2i (3-8i)2i(3−8i)2, i, left parenthesis, 3, minus, 8, i, right parenthesis. Write the resulting number in the form of a+bia+bia, plus, b, i.

Solution

Again, let's start by distributing the 2i2i2, i to each term in the parentheses.

\begin{aligned}\tealD{2i}(3-8i)&=\tealD{2i}(3)-\tealD{2i}(8i)\\ \\ &=6i-16i^2 \end{aligned}

2i(3−8i)

​

=2i(3)−2i(8i)

=6i−16i

2

​

At this point, the answer is not of the form a+bia+bia, plus, b, i since it contains i^2i

2

i, squared.

However, we know that \goldD{i^2=-1}i

2

=−1start color #e07d10, i, squared, equals, minus, 1, end color #e07d10. Let's substitute and see where that gets us.

\begin{aligned}\phantom{\tealD{2i}(3-8i)} &=6i-16\goldD{i^2}\\ \\ &=6i-16(\goldD{-1})\\ \\ &=6i+16\\ \end{aligned}

2i(3−8i)

​

=6i−16i

2

=6i−16(−1)

=6i+16

​

Using the commutative property, we can write the answer as 16+6i16+6i16, plus, 6, i, and so we have that 2i (3-8i)=16+6i2i(3−8i)=16+6i2, i, left parenthesis, 3, minus, 8, i, right parenthesis, equals, 16, plus, 6, i.