Multiplying algebraic fractions with 6x^2+5x-6

[aricho]

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

 

[Fermat]

You're meant to simplify the expression, yes?

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

 

[aricho]

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

 

[thorney]

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

 

[aricho]

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

 

[aricho]

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

 

[bomba923]

Hmm, I seem to get
$$\left( {\frac{{6x^2 + 5x - 6}}<br /> {{6x^2 + 13x + 6}}} \right)\left( {\frac{{3x^2 - 4x - 4}}<br /> {{3x^2 - 8x + 4}}} \right) = \left[ {\frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}}} \right]\left[ {\frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}}} \right] = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x - 2} \right)\left( {2x + 3} \right)}}$$

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

Then I suppose?

$$\frac{{6x^2 + 5x - 6}}{{6x^2 + 13x + 6}} + \frac{{3x^2 - 4x - 4}}<br /> {{3x^2 - 8x + 4}} = \frac{{\left( {3x + 2} \right)\left( {2x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x + 3} \right)}} + \frac{{\left( {3x + 2} \right)\left( {x - 2} \right)}}{{\left( {3x - 2} \right)\left( {x - 2} \right)}} = \frac{{\left( {2x - 3} \right)\left( {3x - 2} \right) + \left( {3x + 2} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} =$$
$$\frac{{2x - 3}}{{2x + 3}} + \frac{{3x + 2}}{{3x - 2}} = \frac{{12\left( {x^2 + 1} \right)}}{{\left( {2x + 3} \right)\left( {3x - 2} \right)}} = \frac{4}{{3x - 2}} - \frac{6}{{2x + 3}} + 2$$

 

[aricho]

is that with a plus in between the fractions?

 

[bomba923]

Yes, LaTex doesn't clearly space out the signs

 

[TD]

I believe it's a but simpler, perhaps you made some slight mistakes.

$$\frac{{6x^2 + 5x - 6}}<br /> {{6x^2 + 13x + 6}} \cdot \frac{{3x^2 - 4x - 4}}<br /> {{3x^2 - 8x + 4}} = \frac{{\left( {2x + 3} \right)\left( {3x - 2} \right)}}<br /> {{\left( {2x + 3} \right)\left( {3x + 2} \right)}} \cdot \frac{{\left( {x - 2} \right)\left( {3x + 2} \right)}}<br /> {{\left( {x - 2} \right)\left( {3x - 2} \right)}} = 1$$

Cancels out nicely

 

[aricho]

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

 

[thorney]

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