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

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me solve this: {{{(2x^2-2)/(6x+6)}}} times {{{(6x^2+18x)/(x^2+2x-3)}}}       Log On


   

Question 78330: Please help me solve this:
%282x%5E2-2%29%2F%286x%2B6%29 times %286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29
Found 2 solutions by jim_thompson5910, rmromero:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28%282x%5E2-2%29%2F%286x%2B6%29%29%2A%28%286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29%29

%28%282x%5E2-2%29%2F%286%28x%2B1%29%29%29%2A%28%286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29%29 Factor a 6 out of 6x%2B6

%282%28x%5E2-1%29%2F%286%28x%2B1%29%29%29%2A%28%286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29%29 Factor a 2 out of 2x%5E2-2

%282%28x%5E2-1%29%2F%286%28x%2B1%29%29%29%2A%286x%28x%2B3%29%2F%28x%5E2%2B2x-3%29%29 Factor a 6x out of 6x%5E2%2B18x

Now factor x%5E2-1

Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B0%2Ax%2B-1, 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: %28x%2Ba%29%28x%2Bb%29substitute 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 1%2Ax%5E2%2B0%2Ax%2B-1 again



Now factor x%5E2%2B2x-3
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B2%2Ax%2B-3, 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: %28x%2Ba%29%28x%2Bb%29substitute 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 1%2Ax%5E2%2B2%2Ax%2B-3 again



So after all of that we get


Notice these terms cancel
So we're left with
%282%2F6%29%2A%286x%2F1%29

%282%2A6x%2F6%29 Multiply

%2812x%2F6%29

2x Divide
So the expression %28%282x%5E2-2%29%2F%286x%2B6%29%29%2A%28%286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29%29 reduces to
2x
Answer by rmromero(383) About Me  (Show Source):
You can put this solution on YOUR website!


Please help me solve this:



%282x%5E2-2%29%2F%286x%2B6%29 times %286x%5E2%2B18x%29%2F%28x%5E2%2B2x-3%29







  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 cross+%28%28x%2B1%29%28x-1%29%29              cross+%286%29x cross%28x%2B3%29

________________   *     ______________

  cross+%286+%28x%2B1%29%29                  cross+%28%28x-1%29%28x%2B3%29%29









         highlight+%282x%29