Factoring Help needed...

[jokerfmj]

Hey, all.

I have been given two expressions that I need to factor and then divide and I am really stuck on this and would appreciate any/all help.

The first is:

10b - 6 / 3b^2 - 12b + 12 (/ is divide by for me... hope that is clear.) I was told I had to factor the denominator and I did the following steps:

3b^2 - 12b + 12=0

3(b^2 - 6b) - 6b + 12=0

3b(b - 2) - 6(b - 2)=0

(b - 2)(3b - 6)=0

b-2=0 or 3b-6=0

b=2 or 3b=6
b=2 or b=2

I'm fairly certain this one is correct. The second expression whose denominator I have to factor is giving me fits though:

The second expression is: (5b - 3) / (3b^2 - 12) = 0

3b^2 - 12 = 0

3(b^2 - 4) = 0

(b - 2)(b + 2) = 0

b - 2 = 0 or b + 2 = 0

I'm not sure about this one... I'm not sure if dividing by 3 is allowed to eliminate it from the b^2-4 portion, although from what I've looked up that seems to be how to do it.

So that is my first question, are they done properly?

Secondly, what is giving me the most trouble, is that I have to divide the first expression by the second expression, which looks like this:

((10b - 6) / (3b^2 - 12b + 12)) / ((5b - 3) / (3b^2 - 12))

So the guidance I am given tells me to replace the denominators with the factored out expressions and then flip the second expression and change the division sign to multiplication. I understand flipping the second expression and changing it to multiplication, but I don't understand if I am supposed to use (b - 2)(b + 2) for the second denominator or 3(b - 2)(b + 2), or 3(b^2 - 4) for it.

Can anyone offer help on this?

Thanks, and sorry if it's confusing.

 

[serds]

hey,

a product is 0 if one of the factors is 0.
so if 3*0=0 , 3*(b - 2)(b + 2) = 0 if b=2 or b=-2.
You have to use 3(b - 2)(b + 2) or 3(b^2 - 4)

 

[jokerfmj]

hey,

a product is 0 if one of the factors is 0.
so if 3*0=0 , 3*(b - 2)(b + 2) = 0 if b=2 or b=-2.
You have to use 3(b - 2)(b + 2) or 3(b^2 - 4)

Gotcha... Ok, so I used this expression:

((10b - 6) / (3b^2 - 12b + 12)) ÷ ((5b-3)(3b^2 - 12))

I then flipped it to this one:

((10b - 6) / (b-2)(3b-6)) ∙ ((3(b-2)(b+2)) / (5b-3))

I eventually ended up with:

(-10b + 6) / (5b - 3) which comes out to -2. (Is that right?)


If that's right, then the next part I'm stuck on is finding the Least Common Denominator so that I can add these two rational expressions...

Can anyone help with that?

 

[jokerfmj]

((10b - 6)/(3b^2 - 12b + 12)) + ((5b - 3)/(3b^2 - 12))

Basically I have to figure out how to build these up using the Least Common Denominator and then add them.

Hopefully that was more clear than my last post.

 

[serds]

3b^2 - 12b + 12 = 3 (b - 2)²
3b^2 - 12 = 3 (b - 2) (b + 2)
LCD = 3 (b - 2)² (b + 2)

 

[jokerfmj]

3b^2 - 12b + 12 = 3 (b - 2)²
3b^2 - 12 = 3 (b - 2) (b + 2)
LCD = 3 (b - 2)² (b + 2)

How do I apply that to the expressions and add them?

I'm not normally this needy when it comes to math, it's just that this if the first week and I was given 1 day to do this assignment, read the book, do the lab, and it's all kind of out of whack. Once I get this weekend to get on track I'll be ok... Any and all help is appreciated on this!

 

[serds]

take each factor of the denominators. each factor has to be in the LCD (with the highest exponent (b - 2)² for example).
^{then multiply the other fraction with the one factor, that was not in the factorised denominator.
here:}
(10b - 6) * (b+2)
(5b - 3) * (b-2)
Then you can add them

(10b - 6) * (b+2) + (5b - 3) * (b-2)
LCD

 

[jokerfmj]

take each factor of the denominators. each factor has to be in the LCD (in the highest potency (b - 2)² for example).
^{then multiply the other fraction with the one factor, that was not in the factorised denominator.
here:}
(10b - 6) * (b+2)
(5b - 3) * (b-2)
Then you can add them

(10b - 6) * (b+2) + (5b - 3) * (b-2)
LCD

Ok, I think I get that now... But what about the denominator? If the numerator gets multiplied by something than the denominator also has to be multiplied by the same to keep the fraction equivalent, correct?

(10b - 6) .....* ... (5b - 3)
3b² - 12b + 12 ... 3b² - 12

I just used the dots to keep things lined up...

So I would multiply the left side numerator by (b+2) and the right side numerator by (b-2), do I multiply the respective denominators by the same? Or does the denominator change to the factored out form, so the denominator would be:

3(b-2)²(b+2)?

 

[serds]

You have already found the least common denominator, so you just take this:

(10b - 6) * (b+2) + (5b - 3) * (b-2)
3 (b - 2)² (b + 2)

Sorry, I'm not used to the correct terms due to English is not my mother tongue.

 

[jokerfmj]

You have already found the least common denominator, so you just take this:

(10b - 6) * (b+2) + (5b - 3) * (b-2)
3 (b - 2)² (b + 2)

Sorry, I'm not used to the correct terms due to English is not my mother tongue.

You have been a great help! I cannot express to you how much I appreciate you taking the time to help me on this.

Thank you very much!

No worries, I understood your English just fine!

 

[serds]

It was a pleasure!