Question

(Algebra 2 Honors)
Complete the following exercises by applying polynomial identities to complex numbers.

1.Factor x2 + 64. Check your work.
2.Factor 16x2 + 49. Check your work.
3.Find the product of (x + 9i)2.
4.Find the product of (x – 2i)2.
5.Find the product of (x + (3+5i))2.

Answer 1

\(x^2+64\) is a perfect square, it factors with complex numbers like this:
\((x+8i)(x-8i)\)
To understand or confirm it you need to know that \(i^2=-1\).
You should be able to do the second one by using exactly the same method. For the last 3 you just need to treat \(i\) like a variable like \(y\).

Answer 2

Um.. Alright, I suppose. So..
16X^2 + 49 would be
(x + 16i)(x + 16i) + 49
x^2 + 32 + 49
x^2 + 81 Worked out??

Answer 3

Not really, \(16x^2=(4x)^2\).
So \((4x)^2+7^2=(4x+7i)(4x-7i)\).

Answer 4

So I'm supposed to get the square root first??

Answer 5

(x + 9i)^2

So..
(x + 3i)(x + 3i)
x^2 + 9i
...I don't think I did that right.

Answer 6

Ah, they were squares. I thought it was just multiply by 2.
Anyhow, just treat \(i\) like a regular variable:
\((x+9i)^2=(x+9i)(x+9i)=x^2+18ix+81i^2\)
Then in the end, replace \(i^2=-1\)
\[=x^2+18ix-81.\]

Answer 7

Okay, do you do anything with it after that, then??

Answer 8

Not really, we have found the product =)

Answer 9

Oh, okay. I'll try doing 4, then.
(x – 2i)^2
(x - 2i)(x - 2i)
x^2 + 4i - 2ix - 2ix
Is that right..?

Answer 10

Yeah, almost. It should just be \(4i^2=-4\) instead of \(4i\).

Answer 11

Oops, okay. So, for number 5..
(x + (3+5i))^2
Um.. How exactly would I go about doing this one? ._.;;;

Answer 12

Just keep thinking of \(i\) like another \(x\) until the end ;)

Answer 13

I mean to do the multiplication/factoring.. How would I go about that with the parenthesis inside the parenthesis??

Answer 14

Oh, either think of that parenthesis as one 'thing' or just ignore them.
So you could do \((x+(3+5i))^2=x^2+x(3+5i)+x(3+5i)+(3+5i)^2\)
or \((x+3+5i)^2=x^2+3x+5xi+\ldots\)etc

Answer 15

Okay, so.. Would you mind checking my work, and make sure I did everything right, please?

X^2 + 64
(x + 8i)(x + 8i)
x^2 + 16i^2 + 16ix^2
x^2 - 16 + 16ix^2

16x^2 + 49
(4x)^2 + 72
(4x + 7i)(4x + 7i)
16x^2 + 49i^2 + 28ix^2
16x^2 - 49 + 28ix^2

(x + 9i)^2
(x + 9i)(x + 9i)
x^2 + 18ix + 81i^2
x^2 + 18ix – 81

(x – 2i)^2
(x – 2i)(x – 2i)
x^2 + 4i^2 – 2ix – 2ix
x^2 – 4 – 2ix – 2ix

(x + (3 + 5i))^2
x^2 + x(3 + 5i) + x(3 + 5i) + (3 + 5i)^2

Answer 16

Yes it's correct, although you shouldn't expand after you have factored in the first two, and it should be one factor with plus, one with minus, like this for the first one:
\[(x+8i)(x-8i)\]
That's factored, so that's the final answer.
Then in the last one you need to expand the remaining parentheses.

Answer 17

Okay, I'll fix those two, then.
How do I factor out the remaining parentheses, in the last one, though..? I'm sorry, I'm really bad at math.

Answer 18

No worries, just do it like you did in the previous ones. Multiply each term with every other term so \(x(3+5i)=3x+5ix\)

Answer 19

...OHH! Do I combine like terms after, 'cause I feel like there's gonna be alot of repeats.

Answer 20

Yeah, it's always a good idea to combine like terms when you see them =)

Answer 21

Okay, cool. :D Gimme a sec, and I'll work on it.

Answer 22

(x + (3 + 5i)^2
x^2 + x(3 + 5i) + x(3 + 5i) + (3 + 5i)^2
x^2 + 3x + 5ix + 3x + 5ix + (3 + 5i)(3 + 5i)
x^2 + 3x + 5ix + 3x + 5ix + 9 + 25i2 + 15i + 15i
x^2 + 3x^2 + 5ix^2 + 9 + 25i^2 + 15i^2
x^2 + 3x^2 + 5ix^2 + 9 - 25 – 15
x^2 + 3x^2 + 5ix^2 - 31

Did I do that right??

Answer 23

No, you combined the 3x's and 5ix's in the wrong way. 3x=x+x+x, so 3x+3x=x+x+x+x+x+x=6x.
Or you can think of it 3x+3x=(3+3)x=6x. Same thing for 5ix+5ix=10ix.

Answer 24

OHHHHH Okay! Sorry about that! ^^;;;;

Answer 25

Thank you for the help! :D I really appreciate it!

Answer 26

You're welcome =D