Question

Simplify. How and Why? ->

Answer 1

?

Answer 2

there :)

Answer 3

okay when u see division
1/2 / 2/3 = 1/2*3/2

Answer 4

so do the flippy thing on your picture

Answer 5

could i use synthetic division here?

Answer 6

(1+3/x)(x-2)4/[(x^2-4)(x^2+5x+6) preformed division
4(x+3)(x-2)/[x(x^2-4)(x+3)(x+2) common denominator and factor
4(x+3)(x-2)/[x(x-2)(x+2)(x+3)(x+2) factored again
4/[x(x+2)^2}
from this you see vertical asymptote at 0 and -2 there vertical asymptote because y is undefined at these points

Answer 7

assuming you set equal to y

Answer 8

\[(1 + {3\over x})({x-2\over x^2 + 5x +6}) \div ({ x^2-4 \over4}) \]First let's look at the right hand part. Let's multiply through so we have one single big term. First factor the quadratic and then multiply out.\[(1 + {3\over x})({x-2\over (x+2)(x+3)})\]\[({x-2\over (x+2)(x+3)}) + ({3(x-2)\over x(x+2)(x+3)})\]Now divide the add them together by modifying the right hand term with x/x\[({x(x-2)\over x(x+2)(x+3)}) + ({3(x-2)\over x(x+2)(x+3)})\]Now the denominators are the same, so we can add them. \[ {x(x-2)+3(x-2)\over x(x+2)(x+3)}\]Noticing the common factor in the numerator (x-2) we can rewrite as\[ {(x+3)(x-2)\over x(x+2)(x+3)}\]Cancel\[ {\cancel{(x+3)}(x-2)\over x(x+2)\cancel{(x+3)}}\]Leaving\[ {(x-2)\over x(x+2)}\]All that is left is the division bit.

\[ {(x-2)\over x(x+2)} \div ({ x^2-4 \over4})\]which can be rewritten as\[ {(x-2)\over x(x+2)} \times ({ 4 \over x^2-4})\]\[ {(x-2)\over x(x+2)} \times ({ 4 \over (x+2)(x-2)})\]\[ {\cancel{(x-2)}\over x(x+2)} \times ({ 4 \over (x+2)\cancel{(x-2)}})\]Leaving \[4\over x(x+2)^2\]

Answer 9

Hope that helps :)

Answer 10

checks

Answer 11

woops, didn't realise it was already done, sorry

Answer 12

nice to have a check!!!

Answer 13

@karenangelahernandez do you get it?

Answer 14

apparently, she gave me metals