Question

What is the restriction on the quotient of quantity 5 x squared minus 10 x minus 15 divided by quantity x squared minus 9 divided by quantity 6 x plus 12 divided by quantity x plus 2?

Answer 1

so ur equation is (5x^2 -10x -15)/(x^2 -9)/(6x+12)/(x^+2)?

Answer 2

Yep, that's it:)

Answer 3

How would I go about solving this?

Answer 4

just by looking at the equation u know the top part can be 0 and all real numbers known as IR. The others can have possible restrictions if SET EQUAL TO ZERO because it is undefined when u a number for example 1/0.

Answer 5

I'm a bit lost, how do I set these to zero?

Answer 6

so really you have (5x^2-10x-15)/(x^2 -9)(6x+12)(x+2)

Answer 7

all u care about is the restriction right. so set those () each = 0 and solve for x

Answer 8

I get 'how' I would set these to zero, but how would I solve the parts with that? Aren't there no like terms.?

Answer 9

well you are dividing by a fraction so flip and multiply which gives

\[\frac{5x^2 -10x -15}{x^2 -9} \times \frac{6x + 12}{x + 2}\]

so to find the restrictions... I'd first factor the quadratics then write everything over a common denominator

\[\frac{5(x +1)(x-3)6(x + 2)}{(x +3)(x-3)(x+2)}\]

so the restrictions occur in the denominator...

identify the values of x that make the denominator zero...

there will be 2 points of discontinuity and 1 vertical asymptote

Answer 10

depending on the way it should be answered the expression above can be simplified to

\[\frac{30(x + 1)}{(x +3)}\]

Answer 11

by stating the restrictions on the simplified form you will only get 1 vertical asymptote...

Answer 12

ok, so I then cancel some of the terms. I was confused by this because they want me to find the restriction of the product.? Would it be -3 since that's the opposite of x + 3?

Answer 13

well there are actually 3 restrictions....

2 are points of discontinuity... there occur in the original product... but not in the simplified form.

there is 1 restriction that occurs in the original and simplified form...

the question doesn't state how many answers ...

but for me... I'd provide all 3 ...

Answer 14

-3 would make the the equation be 0/0 which is ok

Answer 15

x = -3 doesn't but x = 3 does...

Answer 16

when you get 0/0 you get a point of discontinuity

Answer 17

here is a nice summary
http://www.math.brown.edu/UTRA/discontinuities.html

Answer 18

so is it -3 or 3?
I have both options..

Answer 19

plug what u find into the top part of orginal equation because u know the bottom is ?/0

Answer 20

well there are both restictions... its just determining their type... there is 1 other restiction in the domain

Answer 21

If the denominator was x + 3, I would think that it would be the opposite of -3, but I'm not that good. My options are:
x ≠ -2

x ≠ -3

x ≠ 3

x ≠ 9

Answer 22

oops slight typo

should be

\[\frac{5(x +1)(x-3)(x +2)}{(x-3)(x+3)6(x+2)}\]

so cancel the common factors...


and look at the denominator restriction...

Answer 23

x value = -3 plug into your equation check to see if it is undefined

Answer 24

so simplifying you get

\[\frac{5(x +1)}{6(x +3)}\]

the restriction occurs when

x + 3 = 0

hope it helps

Answer 25

Ok, well, thanks, both of you for your help:)

Answer 26

yw just hope ur not more confused and u understand it better.

Answer 27

I do, thx again, :)
I just posted a similar one but is just asking for simplification if you have any time for one more.

Answer 28

so 3 is the restriction?

Answer 29

I think it would be negative three, because the restriction is generally the opposite. Not sure though.

Answer 30

Yep, just finshed the test- the answer is negative 3