Solving Algebraic Fractions: 6x^2+5x-6 & 3x^2-4x-4

  • Thread starter aricho
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In summary, the person is having trouble with factoring a two-term polynomial. They ask for help and provide a summary of the content. They explain that they can factor the expression, but they don't know what to do after that. They eventually get to a solution, but it's not easy.
  • #1
aricho
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hey,
I am having trouble with the following...

(6x^2+5x-6)/(6x^2+13x+6) TIMES (3x^2-4x-4)/(3x^2-8x+4)

i can factorise them but i don't know what to do after that...

Thanks for your help
 
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  • #2
You're meant to simplify the expression, yes?

Just cancel out the common factors. I don't think it can be simplified any further
 
  • #3
yer, but what is left is nasty haha and its wrong
 
  • #4
it simplifys all the way too 1 eventually doesn't it?
 
  • #5
ummm nar...

this is what i can get to...
{(3x+3)(2x-2)/(2x+3)(3x+2)} + {(3x+2)(x-2)/(3x-2)(x-2)}

and the answer is...

{2(9x^2+4)}/(3x-2)(3x+2)

i just don't know how to get there haha
 
  • #6
nice...it does...how about it with a plus instead of a multiply between the two factions
 
  • #7
Hmm, I seem to get
[tex]\left( {\frac{{6x^2 + 5x - 6}}
{{6x^2 + 13x + 6}}} \right)\left( {\frac{{3x^2 - 4x - 4}}
{{3x^2 - 8x + 4}}} \right) = \left[ {\frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}}} \right]\left[ {\frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}}} \right] = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x - 2} \right)\left( {2x + 3} \right)}} [/tex]

aricho said:
nice...it does...how about it with a plus instead of a multiply between the two factions
Then I suppose?

[tex]\frac{{6x^2 + 5x - 6}}{{6x^2 + 13x + 6}} + \frac{{3x^2 - 4x - 4}}
{{3x^2 - 8x + 4}} = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}} + \frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}} = \frac{{\left( {2x - 3} \right)\left( {3x - 2} \right) + \left( {3x + 2} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} = [/tex]
[tex] \frac{{2x - 3}}{{2x + 3}} + \frac{{3x + 2}}{{3x - 2}} = \frac{{12\left( {x^2 + 1} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} = \frac{4}{{3x - 2}} - \frac{6}{{2x + 3}} + 2 [/tex]
 
Last edited:
  • #8
is that with a plus in between the fractions?
 
  • #9
Yes, LaTex doesn't clearly space out the signs :rolleyes:
 
  • #10
I believe it's a but simpler, perhaps you made some slight mistakes.

[tex]\frac{{6x^2 + 5x - 6}}
{{6x^2 + 13x + 6}} \cdot \frac{{3x^2 - 4x - 4}}
{{3x^2 - 8x + 4}} = \frac{{\left( {2x + 3} \right)\left( {3x - 2} \right)}}
{{\left( {2x + 3} \right)\left( {3x + 2} \right)}} \cdot \frac{{\left( {x - 2} \right)\left( {3x + 2} \right)}}
{{\left( {x - 2} \right)\left( {3x - 2} \right)}} = 1[/tex]

Cancels out nicely :smile:
 
  • #11
hmmm, maybe the answer is wrong... it says {2(9x^2+4)}/(3x-2)(3x+2)
 
  • #12
or for the multiplication one you can expand the polynomial and the top clearly equals the bottom with reduced thinking needed!
 
Last edited:

1. How do I simplify this algebraic fraction?

To simplify an algebraic fraction, you need to factor both the numerator and the denominator. Then, you can cancel out any common factors and rewrite the fraction in its simplest form.

2. Can I solve for x in this expression?

Yes, you can solve for x by setting the fraction equal to a number and then solving for x using algebraic techniques such as factoring, completing the square, or using the quadratic formula.

3. What is the degree of this algebraic fraction?

The degree of an algebraic fraction is determined by the highest exponent of the variable present in either the numerator or denominator. In this case, both fractions have a degree of 2.

4. Can I simplify this fraction further?

It depends on the given expression. Some algebraic fractions can be simplified to a point where they cannot be simplified any further. If the numerator and denominator have no common factors, then the fraction is already in its simplest form.

5. Are there any restrictions on the values of x in this expression?

Some algebraic fractions may have restrictions on the values of x in order for the fraction to be defined. For example, if the denominator has a variable in the form of a square root, the value of x must be non-negative in order for the fraction to be defined. It is important to check for any restrictions when solving algebraic fractions.

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