SOLUTION: Please help me solve. thanks! (w^2+2w)/(w^2-9) divided by (w^2+7w+10)/(w^2+8w+15)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me solve. thanks! (w^2+2w)/(w^2-9) divided by (w^2+7w+10)/(w^2+8w+15)      Log On


   

Question 77107: Please help me solve. thanks!
(w^2+2w)/(w^2-9) divided by (w^2+7w+10)/(w^2+8w+15)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28%28w%5E2%2B2w%29%2F%28w%5E2-9%29%29%2F%28%28w%5E2%2B7w%2B10%29%2F%28w%5E2%2B8w%2B15%29%29



%28w%28w%2B2%29%2F%28w%5E2-9%29%29%2F%28%28w%5E2%2B7w%2B10%29%2F%28w%5E2%2B8w%2B15%29%29 Factor w out of the numerator
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Factor the denominator %28w%5E2-9%29
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B0%2Ax%2B-9, first we need to ask ourselves: What two numbers multiply to -9 and add to 0? Lets find out by listing all of the possible factors of -9


Factors:

1,3,9,

-1,-3,-9,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -9.

(-1)*(9)=-9

(-3)*(3)=-9

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

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First Number|Second Number|Sum
1|-9|1+(-9)=-8
3|-3|3+(-3)=0
-1|9|(-1)+9=8
-3|3|(-3)+3=0
We can see from the table that -3 and 3 add to 0.So the two numbers that multiply to -9 and add to 0 are: -3 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=-3 and b=3 So the equation becomes: (x-3)(x+3) Notice that if we foil (x-3)(x+3) we get the quadratic 1%2Ax%5E2%2B0%2Ax%2B-9 again


So the denominator %28w%5E2-9%29 factors to:
%28w-3%29%28w%2B3%29
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Factor the numerator %28w%5E2%2B7w%2B10%29
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B7%2Ax%2B10, first we need to ask ourselves: What two numbers multiply to 10 and add to 7? Lets find out by listing all of the possible factors of 10


Factors:

1,2,5,10,

-1,-2,-5,-10,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 10.

1*10=10

2*5=10

(-1)*(-10)=10

(-2)*(-5)=10

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7

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First Number|Second Number|Sum
1|10|1+10=11
2|5|2+5=7
-1|-10|-1+(-10)=-11
-2|-5|-2+(-5)=-7
We can see from the table that 2 and 5 add to 7. So the two numbers that multiply to 10 and add to 7 are: 2 and 5 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=2 and b=5 So the equation becomes: (x+2)(x+5) Notice that if we foil (x+2)(x+5) we get the quadratic 1%2Ax%5E2%2B7%2Ax%2B10 again



So the numerator w%5E2%2B7w%2B10 factors to:
%28w%2B2%29%28w%2B5%29
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Factor the denominator %28w%5E2%2B8w%2B15%29
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B8%2Ax%2B15, first we need to ask ourselves: What two numbers multiply to 15 and add to 8? Lets find out by listing all of the possible factors of 15


Factors:

1,3,5,15,

-1,-3,-5,-15,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 15.

1*15=15

3*5=15

(-1)*(-15)=15

(-3)*(-5)=15

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8

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First Number|Second Number|Sum
1|15|1+15=16
3|5|3+5=8
-1|-15|-1+(-15)=-16
-3|-5|-3+(-5)=-8
We can see from the table that 3 and 5 add to 8. So the two numbers that multiply to 15 and add to 8 are: 3 and 5 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=3 and b=5 So the equation becomes: (x+3)(x+5) Notice that if we foil (x+3)(x+5) we get the quadratic 1%2Ax%5E2%2B8%2Ax%2B15 again



So the denominator w%5E2%2B8w%2B15 factors to:
%28w%2B3%29%28w%2B5%29
So the whole expression becomes






Which reduces to:
w%2F%28w-3%29