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Algebraic Fraction

  • Thread starter aricho
  • Start date
71
0
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 dont know what to do after that...

Thanks for your help
 

Answers and Replies

Fermat
Homework Helper
872
1
You're meant to simplify the expression, yes?

Just cancel out the common factors. I don't think it can be simplified any further
 
71
0
yer, but what is left is nasty haha and its wrong
 
4
0
it simplifys all the way too 1 eventually doesn't it?
 
71
0
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 dont know how to get there haha
 
71
0
nice...it does....how about it with a plus instead of a multiply between the two factions
 
732
0
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:
71
0
is that with a plus in between the fractions?
 
732
0
Yes, LaTex doesn't clearly space out the signs :rolleyes:
 
TD
Homework Helper
1,020
0
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:
 
71
0
hmmm, maybe the answer is wrong.... it says {2(9x^2+4)}/(3x-2)(3x+2)
 
4
0
or for the multiplication one you can expand the polynomial and the top clearly equals the bottom with reduced thinking needed!
 
Last edited:

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