- #1

- 71

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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

- Thread starter aricho
- Start date

- #1

- 71

- 0

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

- #2

Fermat

Homework Helper

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Just cancel out the common factors. I don't think it can be simplified any further

- #3

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yer, but what is left is nasty haha and its wrong

- #4

- 4

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it simplifys all the way too 1 eventually doesn't it?

- #5

- 71

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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

- #6

- 71

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nice...it does....how about it with a plus instead of a multiply between the two factions

- #7

- 759

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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]

[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]

[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]

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

[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

- 71

- 0

is that with a plus in between the fractions?

- #9

- 759

- 0

Yes, LaTex doesn't clearly space out the signs

- #10

TD

Homework Helper

- 1,022

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[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

- #11

- 71

- 0

hmmm, maybe the answer is wrong.... it says {2(9x^2+4)}/(3x-2)(3x+2)

- #12

- 4

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or for the multiplication one you can expand the polynomial and the top clearly equals the bottom with reduced thinking needed!

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