Question

Help with a restriction of a product and quotient question. Just need some help. Will award best answer

Answer 1

@iambatman

Answer 2

@e.mccormick

Answer 3

@jim_thompson5910

Answer 4

are you able to factor each piece?

Answer 5

yes, I can

Answer 6

(x+4)(x+1) / (x+4)(x+1) * (x+5)(x-1) /(x+3)
Those are the factors I found.
I want to learn how to find restrictions, can you also help and teach me do that? I want to learn this and understand this! @jim_thompson5910

Answer 7

you factored x^2 + 3x - 4 incorrectly

Answer 8

everything else looks good, so try factoring x^2 + 3x - 4 again

Answer 9

oops sorry, okay

Answer 10

It should be (x-1) instead of (x+1) my typing error

Answer 11

@jim_thompson5910

Answer 12

so we have

\[\Large \frac{x^2+5x+4}{x^2+3x-4} \cdot \frac{x^2+4x-5}{x+3}\]
\[\Large \frac{(x+4)(x+1)}{(x+4)(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]

Answer 13

looking at
\[\Large \frac{(x+4)(x+1)}{(x+4)(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]
what do you see canceling?

Answer 14

(x+4)

Answer 15

\[\Large \frac{(x+4)(x+1)}{(x+4)(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{\cancel{(x+4)}(x+1)}{\cancel{(x+4)}(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{x+1}{x-1} \cdot \frac{(x+5)(x-1)}{x+3}\]

Answer 16

(x-1)

Answer 17

\[\Large \frac{x+1}{x-1} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{x+1}{\cancel{(x-1)}} \cdot \frac{(x+5)\cancel{(x-1)}}{x+3}\]
\[\Large \frac{x+1}{1} \cdot \frac{x+5}{x+3}\]
\[\Large \frac{(x+1)(x+5)}{1*(x+3)}\]
\[\Large \frac{(x+1)(x+5)}{x+3}\]
\[\Large \frac{x^2+6x+5}{x+3}\]

Answer 18

so this is what the full step by step picture looks like

\[\Large \frac{x^2+5x+4}{x^2+3x-4} \cdot \frac{x^2+4x-5}{x+3}\]
\[\Large \frac{(x+4)(x+1)}{(x+4)(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{\cancel{(x+4)}(x+1)}{\cancel{(x+4)}(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{x+1}{x-1} \cdot \frac{(x+5)(x-1)}{x+3}\]
\[\Large \frac{x+1}{\cancel{(x-1)}} \cdot \frac{(x+5)\cancel{(x-1)}}{x+3}\]
\[\Large \frac{x+1}{1} \cdot \frac{x+5}{x+3}\]
\[\Large \frac{(x+1)(x+5)}{1*(x+3)}\]
\[\Large \frac{(x+1)(x+5)}{x+3}\]
\[\Large \frac{x^2+6x+5}{x+3}\]

Answer 19

(x+5)(x+1) / x+3

Answer 20

would the restriction be 1?
@jim_thompson5910

Answer 21

the restriction is basically kicking out numbers that make any denominator zero

Answer 22

so one number we kick out is x = -3 because it makes the denominator "x+3" equal to zero

Answer 23

Ohhhh, now that makes sense

Answer 24

@jim_thompson5910

Answer 25

so would 1 be the restriction?

Answer 26

focus on
\[\Large \frac{(x+4)(x+1)}{(x+4)(x-1)} \cdot \frac{(x+5)(x-1)}{x+3}\]

Answer 27

the denominators are: x+4, x-1, x+3

Answer 28

so there are 3 restrictions

one of the three restrictions are x = -3 (it makes x+3 equal to zero)

Answer 29

another is x = 1 since it makes x-1 equal to zero

Answer 30

oh would 4 be it, I forgot about the other denominators.

Answer 31

not 4

Answer 32

OHHH, it is what will make it 0... sorry I was confused, so I have 3 answer choices for my study guide and it includes -1, -3 and 4... I do not know what to do. @jim_thompson5910

Answer 33

you can only choose one choice?

Answer 34

Yes... and I have no idea what to do now

Answer 35

ok so why not just look at the final simplified result

Answer 36

there is only one denominator there

Answer 37

Okay! -3 then!

Answer 38

yes

Answer 39

Thank you so much! now I can do this for my next equation with no problem!! thank you!!

Answer 40

you're welcome