Question

Just wanted to check my answers if they are right. (See below)

Answer 1

1.)\[\frac{ x-4 }{ x^2-3x-4 }\]

Answer 2

My answer: \[\frac{ 1 }{ (x+1) }\]

Answer 3

1 is correct

Answer 4

Thanks!
2.) \[\frac{ x^3-8 }{ x-2 }\]

Answer 5

My answer:\[(x+2)^2\]

Answer 6

I think u can simplify more than that

Answer 7

How would I simplify that?

Answer 8

hold on

Answer 9

something is wrong

Answer 10

I don't think I factored the numerator right.

Answer 11

start by putting the 8 into 2^3

Answer 12

Okay. Then what?

Answer 13

and use the difference of cubes which is \[a^3−b^3=(a−b)(a^2+ab+b^2)\]

Answer 14

it will get you \[(x−2)(x^2+(x)(2)+2^2)\]

Answer 15

Simplify 2^2 to 4 and regroup terms

Answer 16

\[\frac{ (x−2)(x^2+2x+4) }{ x-2 }\]

Answer 17

now just cancel the x-2

Answer 18

did you get it?

Answer 19

Ya I got it. Wasn't that the same answer I got?

Answer 20

\[(x+2)^2\]

Answer 21

It's different

Answer 22

I wrote the same answer.

Answer 23

\[x^2+2x +4 \neq (x+2)^2\]

Answer 24

Ya sorry about that. Just noticed.

Answer 25

I would have to use the quadratic formula for this right?

Answer 26

\(a^3-b^3=(x-a)(a^2+ab+b^2)\)
\(x^3-2^3=(x-2)(x^2+2x+2^2)=x^2+2x+4\)

\(\Large\frac{x^3-8}{x-2}=\frac{x^2-2^3}{x-2}=\frac{(x-2)(x^2+2x+4)}{x-2}\)

Answer 27

\(\Large =x^2+2x+4\)

Answer 28

Ya I got that so far. Thanks!

Answer 29

\[\frac{ -2\pm \sqrt{2^2-4(1)(4)} }{ 2(1) }\]\[\frac{ -2\pm \sqrt{4-16} }{ 2 }\]\[\frac{ -2\pm \sqrt{-12} }{ 2 }\]

Answer 30

The answer is going to have an i?

Answer 31

Yes. That's why it cant be factored. If you get complex numbers when factoring, then the polynomial is irreducible. x^2+2x+4 is a prime polynomial.

Answer 32

So I don't have to simplify that then? Do I just put x^2+2x+4?

Answer 33

Yes :)

Answer 34

Alright thanks!
3.)\[\frac{ 5-x }{ x^2-25 }\]

Answer 35

My answer:\[\frac{ -1 }{ x+5 }\]

Answer 36

4.) \[\frac{ x^2-4x-32 }{ x^2-16 }\]

Answer 37

3 is correct

Answer 38

My answer:\[\frac{ x-8 }{ x-4 }\]

Answer 39

Thanks! What about #4?

Answer 40

yes. well done

Answer 41

Thanks you guys!