Question

What is the restriction on the product of x^2 + 3x = 2 / x^2 - 2x - 3 multiplied by x^2 + 4x + 3 / x + 2

Answer 1

that equals sign in there doesn't make much sense, can you double check and rewrite?

Answer 2

use parenthesis

Answer 3

Sorry, it's \[ \frac{ x^2 + 3x + 2 }{ x^2 - 2x - 3 } \times \frac{ x^2 + 4x + 3 }{ x + 2 }\]

Answer 4

okk better. So, can you simplify at all to start?

Answer 5

The numerator of the first fraction can be simplified into (x + 2)(x + 1), the denominator can be simplified into (x - 3)(x + 1), the numerator of the second fraction can be simplified into (x + 3)(x + 1)

Answer 6

good! so you can cancel like polynomials out, what does that leave you with?

Answer 7

It leaves me with the (x + 2), (x - 3), (x + 3), and (x + 1) right?

Answer 8

I'm thinking that the restrictions on the product would stem from values of x that cause either denominator of the original expression to be 0. (Division by zero is undefined.)

The first denominator factors to: ( x + 1) and (x - 3)
The second denominator is just: (x + 2)

To get the restricted values of x, solve each of the following for x:

(x + 1) = 0 (x - 3) = 0 (x + 2) = 0

@maciforster

Answer 9

@Directrix , restriction usually means make it so we dont have an indeterminant or undefined form

Answer 10

and it should still leave you with an equation. Not a listing of binomials

Answer 11

\[\frac{ (x+2)(x+1) }{ (x-3)(x+1) } * \frac{ (x+3)(x+1) }{ (x+2) } = \frac{ (x+2)(x+1)(x+3)(x+1) }{ (x-3)(x+1)(x+2) }\]

As you can see in the right hand side of above, you are able to cancel out (x+2) and (x+1) from both numerator and denominator.
When you cancel a quantity from a fraction or equation, you need to keep track of the RESTRICTION on the variable. In this case both the quantities (x+2) and (x+1) can not be zero.
Sometimes when you solve equations, quantities cancel out, and you need to keep track of that in your final answer.

Answer 12

For instance, if you have:

\[\frac{ (6x+6)(x+3) }{ (x+3) } > 35\]

After canceling out the (x+3), you are left with:

\[6x + 6 > 35\]
-AND-
\[(x+3) \neq 0\]

The second is your restriction. So the final answer would be:

\[x > \frac{ 29 }{ 6 } ,and; x \neq -3\]

Answer 13

in that case the restriction is not part of the solution, so it is not needed.
If x<29/6 then you would need to include the \[x \neq -3\] in your solution as a restriction.

Answer 14

Thank you!!!

Answer 15

@FibonacciChick666

Read the sentence above the "listing of binomials."

To get the restricted values of x, solve each of the following for x:
(x + 1) = 0 (x - 3) = 0 (x + 2) = 0

The set of linear equations is for the Asker to solve to find the restricted values of x.

x = -1, x = 3, and x = -2 are restricted values.

Answer 16

I'm sorry, I read numerator like 12 times