SOLUTION: Please help me solve this: {{{(2x^2-2)/(6x+6)}}} times {{{(6x^2+18x)/(x^2+2x-3)}}}

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Question 78330: Please help me solve this:
times

Found 2 solutions by jim_thompson5910, rmromero:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!


Factor a 6 out of

Factor a 2 out of

Factor a 6x out of

Now factor

Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor , first we need to ask ourselves: What two numbers multiply to -1 and add to 0? Lets find out by listing all of the possible factors of -1


Factors:

1,

-1,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -1.

(-1)*(1)=-1

Now which of these pairs add to 0? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 0

||
First Number|Second Number|Sum
1|-1|1+(-1)=0
-1|1|(-1)+1=0
We can see from the table that -1 and 1 add to 0.So the two numbers that multiply to -1 and add to 0 are: -1 and 1 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: substitute a=-1 and b=1 So the equation becomes: (x-1)(x+1) Notice that if we foil (x-1)(x+1) we get the quadratic again



Now factor
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor , first we need to ask ourselves: What two numbers multiply to -3 and add to 2? Lets find out by listing all of the possible factors of -3


Factors:

1,3,

-1,-3,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -3.

(-1)*(3)=-3

Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2

||
First Number|Second Number|Sum
1|-3|1+(-3)=-2
-1|3|(-1)+3=2
We can see from the table that -1 and 3 add to 2.So the two numbers that multiply to -3 and add to 2 are: -1 and 3 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: substitute a=-1 and b=3 So the equation becomes: (x-1)(x+3) Notice that if we foil (x-1)(x+3) we get the quadratic again



So after all of that we get


Notice these terms cancel
So we're left with


Multiply



Divide
So the expression reduces to

Answer by rmromero(383)   (Show Source): You can put this solution on YOUR website!


Please help me solve this:



 times 







  2x^2 - 2                6x^2 + 18x 

_____________     *     _______________

  6x + 6                 x^2 + 2x - 3





Factor both numerators and denominators if possible



   2(x+1)(x-1)              6x(x+3)

________________   *     ______________

  6 (x+1)                  (x-1)(x+3)



Divide common factors



   2               x 

________________   *     ______________

                    









         

 

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