Simplify the rational expression: (2x^2-12x+18) / (2x^2-18)

[flora33]

(2x^2-12x+18) / (2x^2-18)
[2(x-3)(x-3)] / [2(x^2-9)]

Do I cancel out x-3 twice from the numerator to get:
(2) / [2(x^2-9)]

Then is the 2 in the numerator and denominator also canceled to end with an answer:
x^2-9 ???

I see that 2(x-3)(x-3) could equal 2(x^2-9), which is the monomial in the denominator, but I'm not sure if I should have factored out the denominator to2(x-3)(x-3), so my answer would be 1?

Any help is appreciated1
Flora

 

[stapel]

(2x^2-12x+18) / (2x^2-18) = [2(x-3)(x-3)] / [2(x^2-9)]

Do I cancel out x-3 twice from the numerator to get: (2) / [2(x^2-9)]

Um... I'm sorry, but I don't understand what you mean by "cancelling out" here...? You appear to have made them disappear...? :shock:

"Cancellation" is of course a form of division, using the fact that anything divided by itself is just "1". And a factor of 1 can be ignored in multiplication. This is how 2/6, for instance, simplifies to 1/3: the numerator and denominator are factored to get [(1)(2)]/[(2)(3)], and the 2/2 "cancels" to leave 1/3.

But what division of like by like did you use to make the two factors in the numerator disappear? The denominator looks unchanged to me...?

Please be complete. Thank you!

Eliz.

 

[flora33]

Re:

Um... I'm sorry, but I don't understand what you mean by "cancelling out" here...? You appear to have made them disappear...?

I meant to "factor" them out because they are the same... Sorry, I'm having a hard time with this type of problem, and I'm a little lost!

Flora

 

[Loren]

(2x^2-12x+18) / (2x^2-18)

\(\displaystyle \frac{2(x-3)(x-3)}{2(x^2-9)}\)

Factor the denominator completely, then put to work what Eliz tells you in the preceding post.

 

[flora33]

(2x^2-12x+18) / (2x^2-18)

\(\displaystyle \frac{2(x-3)(x-3)}{2(x^2-9)}\)

Factor the denominator completely, then put to work what Eliz tells you in the preceding post.

\(\displaystyle \frac{2(x-3)(x-3)}{2(x^2-9)}\) would become

\(\displaystyle \frac{2(x-3)(x-3)}{2(x-3)(x-3)}\)

then dividing 2(x-3)(x-3) by 2(x-3)(x-3) I arrive at 1. Am I on the right track here?

Flora

 

[Mrspi]

\(\displaystyle \frac{2(x-3)(x-3)}{2(x^2-9)}\) would become

\(\displaystyle \frac{2(x-3)(x-3)}{2(x-3)(x-3)}\)

then dividing 2(x-3)(x-3) by 2(x-3)(x-3) I arrive at 1. Am I on the right track here?

You have factored the numerator correctly, as 2(x - 3)(x - 3)

Now...the denominator was DIFFERENT, wasn't it? Does it make sense that it factors into exactly the same factors?

Please review the special factoring patterns you should have learned to see if one of them has the same form as x^2 - 9. Here's a website you might want to look at:

http://www.purplemath.com/modules/specfact.htm

Once you've reviewed, give your problem another try.

 

[flora33]

You have factored the numerator correctly, as 2(x - 3)(x - 3)

Now...the denominator was DIFFERENT, wasn't it? Does it make sense that it factors into exactly the same factors?

Please review the special factoring patterns you should have learned to see if one of them has the same form as x^2 - 9. Here's a website you might want to look at:

http://www.purplemath.com/modules/specfact.htm

Once you've reviewed, give your problem another try.

All right, so it should be:
\(\displaystyle \frac{2(x-3)(x-3)}{2(x-3)(x+3)}\)

and my answer will be:
\(\displaystyle \frac{2(x-3)}{2(x+3)}\) or just: \(\displaystyle \frac{(x-3)}{(x+3)}\) ??

 

[Loren]

Looks good to me.