SOLUTION: Please show me how to SOLVE my homework question. thanks.
Y^2-36/y^2-4 multiplied by y^2+6y+8/y^2-2y-24
the choices are:
A.) y+6/y-2 B.) y-6/y+2 c.) y+2/y+6 D.) y-2/y
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Please show me how to SOLVE my homework question. thanks.
Y^2-36/y^2-4 multiplied by y^2+6y+8/y^2-2y-24
the choices are:
A.) y+6/y-2 B.) y-6/y+2 c.) y+2/y+6 D.) y-2/y
Log On
In order to factor , first we need to ask ourselves: What two numbers multiply to -36 and add to 0? Lets find out by listing all of the possible factors of -36
Factors:
1,2,3,4,6,9,12,18,36,
-1,-2,-3,-4,-6,-9,-12,-18,-36,List the negative factors as well. This will allow us to find all possible combinations
These factors pair up to multiply to -36.
(-1)*(36)=-36
(-2)*(18)=-36
(-3)*(12)=-36
(-4)*(9)=-36
(-6)*(6)=-36
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
|
-36
|
|
1+(-36)=-35
2
|
-18
|
|
2+(-18)=-16
3
|
-12
|
|
3+(-12)=-9
4
|
-9
|
|
4+(-9)=-5
6
|
-6
|
|
6+(-6)=0
-1
|
36
|
|
(-1)+36=35
-2
|
18
|
|
(-2)+18=16
-3
|
12
|
|
(-3)+12=9
-4
|
9
|
|
(-4)+9=5
-6
|
6
|
|
(-6)+6=0
We can see from the table that -6 and 6 add to 0.So the two numbers that multiply to -36 and add to 0 are: -6 and 6
Now we substitute these numbers into a and b of the general equation of a product of linear factors which is:
substitute a=-6 and b=6
So the equation becomes:
(x-6)(x+6)
Notice that if we foil (x-6)(x+6) we get the quadratic again
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Factor the first denominator
In order to factor , first we need to ask ourselves: What two numbers multiply to -4 and add to 0? Lets find out by listing all of the possible factors of -4
Factors:
1,2,4,4,6,9,12,18,36,
-1,-2,-4,-4,-6,-9,-12,-18,-36,List the negative factors as well. This will allow us to find all possible combinations
These factors pair up to multiply to -4.
(-1)*(36)=-4
(-2)*(18)=-4
(-4)*(12)=-4
(-4)*(9)=-4
(-6)*(6)=-4
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
|
-36
|
|
1+(-36)=-35
2
|
-18
|
|
2+(-18)=-16
4
|
-12
|
|
4+(-12)=-8
4
|
-9
|
|
4+(-9)=-5
6
|
-6
|
|
6+(-6)=0
-1
|
36
|
|
(-1)+36=35
-2
|
18
|
|
(-2)+18=16
-4
|
12
|
|
(-4)+12=8
-4
|
9
|
|
(-4)+9=5
-6
|
6
|
|
(-6)+6=0
We can see from the table that -6 and 6 add to 0.So the two numbers that multiply to -4 and add to 0 are: -6 and 6
Now we substitute these numbers into a and b of the general equation of a product of linear factors which is:
substitute a=-6 and b=6
So the equation becomes:
(x-6)(x+6)
Notice that if we foil (x-6)(x+6) we get the quadratic again
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Factor the second numerator
In order to factor , first we need to ask ourselves: What two numbers multiply to 8 and add to 6? Lets find out by listing all of the possible factors of 8
Factors:
1,2,4,8,6,9,12,18,36,
-1,-2,-4,-8,-6,-9,-12,-18,-36,List the negative factors as well. This will allow us to find all possible combinations
These factors pair up to multiply to 8.
1*36=8
2*18=8
4*12=8
8*9=8
6*6=8
(-1)*(-36)=8
(-2)*(-18)=8
(-4)*(-12)=8
(-8)*(-9)=8
(-6)*(-6)=8
note: remember two negative numbers multiplied together make a positive number
Now which of these pairs add to 6? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 6
First Number
|
Second Number
|
Sum
1
|
36
|
|
1+36=37
2
|
18
|
|
2+18=20
4
|
12
|
|
4+12=16
8
|
9
|
|
8+9=17
6
|
6
|
|
6+6=12
-1
|
-36
|
|
-1+(-36)=-37
-2
|
-18
|
|
-2+(-18)=-20
-4
|
-12
|
|
-4+(-12)=-16
-8
|
-9
|
|
-8+(-9)=-17
-6
|
-6
|
|
-6+(-6)=-12
substitute a=-6 and b=6
So the equation becomes:
(x-6)(x+6)
Notice that if we foil (x-6)(x+6) we get the quadratic again
None of these factors add to 6. So this quadratic cannot be factored. In order to solve for x, we need to use the quadratic formula.
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Factor the second denominator
In order to factor , first we need to ask ourselves: What two numbers multiply to -24 and add to -2? Lets find out by listing all of the possible factors of -24
Factors:
1,2,3,4,6,8,12,24,36,
-1,-2,-3,-4,-6,-8,-12,-24,-36,List the negative factors as well. This will allow us to find all possible combinations
These factors pair up to multiply to -24.
(-1)*(36)=-24
(-2)*(24)=-24
(-3)*(12)=-24
(-4)*(8)=-24
(-6)*(6)=-24
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
|
-36
|
|
1+(-36)=-35
2
|
-24
|
|
2+(-24)=-22
3
|
-12
|
|
3+(-12)=-9
4
|
-8
|
|
4+(-8)=-4
6
|
-6
|
|
6+(-6)=0
-1
|
36
|
|
(-1)+36=35
-2
|
24
|
|
(-2)+24=22
-3
|
12
|
|
(-3)+12=9
-4
|
8
|
|
(-4)+8=4
-6
|
6
|
|
(-6)+6=0
substitute a=-6 and b=6
So the equation becomes:
(x-6)(x+6)
Notice that if we foil (x-6)(x+6) we get the quadratic again
None of these factors add to -2. So this quadratic cannot be factored. In order to solve for x, we need to use the quadratic formula.
So the whole expression becomes
Cancel like terms
So the whole expression reduces to
So the answer is A)
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