what steps do I take to work this problem? Simplify. 9y^2-3y-2/6y^2-13y-5 divide by 3y^2+10y-8/2y^2+13y+20
arrgghh!!!
arrgghh!!!
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Okay, so to go through it step by step, write the two fractions out on a piece of paper. Now remember, when you divide two fractions, you flip the second fraction (so 1/3 becomes 3/1) and then you multiply them. So (1/2) divided by (1/3) is (1/2) x (3/1). Then multiply the numerators and denominators. So this is (1x3/2 x 1) = 3/2. Got that?
So with your giant equation, flip the second one, and now they can be multipled. But if you factor the quadratic equations first, stuff might cancel out.
So lets start with the numerator of your first 'term': 9y^2-3y-2. This simplifies to: (3y-1)(3y-2). Multiply is out so see, but that should equal 9y^2-3y-2. Now lets do the same to the denominator of that 'term':
6y^2-13y-5 factors to: (3y+1)(2y-5). so your first 'term' is:
[(3y-1)(3y-2)]/ [(3y+1)(2y-5)]
Now your second term. Remember, the numerator, since you flipped it, is: 2y^2+13y+20
Factored, this is: (2y+5)(y+4).
The denominator: 3y^2+10y-8 factors to: (3y-2)(y+4). Now let's put the second term numerator on top of the denominator to get: [(2y+5)(y+4)]/[(3y-2)(y+4)].
You can see right away that the y+4 terms in the numerator and denominator of the second term will cancer out.
Now put the two terms together, and see what else can cancel.
Hope that helps!
So with your giant equation, flip the second one, and now they can be multipled. But if you factor the quadratic equations first, stuff might cancel out.
So lets start with the numerator of your first 'term': 9y^2-3y-2. This simplifies to: (3y-1)(3y-2). Multiply is out so see, but that should equal 9y^2-3y-2. Now lets do the same to the denominator of that 'term':
6y^2-13y-5 factors to: (3y+1)(2y-5). so your first 'term' is:
[(3y-1)(3y-2)]/ [(3y+1)(2y-5)]
Now your second term. Remember, the numerator, since you flipped it, is: 2y^2+13y+20
Factored, this is: (2y+5)(y+4).
The denominator: 3y^2+10y-8 factors to: (3y-2)(y+4). Now let's put the second term numerator on top of the denominator to get: [(2y+5)(y+4)]/[(3y-2)(y+4)].
You can see right away that the y+4 terms in the numerator and denominator of the second term will cancer out.
Now put the two terms together, and see what else can cancel.
Hope that helps!
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[(9y^2-3y-2)/(6y^2-13y+5)]* [(2y^2+13y+20)/(3y^2+10y-8)]
= [(9y^2-6y + 3y -2)/(6y^2-13y+5)]* [(2y^2+8y+5y +20)/(3y^2+12y-2y -8)]
= [3y(3y-2 )+ (3y -2)}/3y(2y-1)-5(2y-1)}]* [2y(y+4)+5(y +4)/(3y(y+4) -2(y+4)]
= {(3y+1)(3y-2)(2y+5)(y+4)]/ {(3y-5)(2y-1)(3y-2)(y+4)}
= {(3y+1)(2y+5)]/ {(3y-5)(2y-1)}.......................Ans
= [(9y^2-6y + 3y -2)/(6y^2-13y+5)]* [(2y^2+8y+5y +20)/(3y^2+12y-2y -8)]
= [3y(3y-2 )+ (3y -2)}/3y(2y-1)-5(2y-1)}]* [2y(y+4)+5(y +4)/(3y(y+4) -2(y+4)]
= {(3y+1)(3y-2)(2y+5)(y+4)]/ {(3y-5)(2y-1)(3y-2)(y+4)}
= {(3y+1)(2y+5)]/ {(3y-5)(2y-1)}.......................Ans
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9*y^2 - 3*y - 2 = (3*y + 1)*(3*y - 2)
6*y^2 - 13*y - 5 = (3*y + 1)*(2*y - 5)
3*y^2 + 10*y - 8 = (y + 4)*(3*y - 2)
2*y^2 + 13*y + 20 = (y + 4)*(2*y + 5)
(3*y + 1)*(3*y - 2)/(3*y + 1)*(2*y - 5) = (3*y - 2)/(2*y - 5), answer
(y + 4)*(3*y - 2)/(y + 4)*(2*y + 5) = (3*y - 2)/(2*y + 5)
(3*y - 2)/(2*y - 5)/(3*y - 2)/(2*y + 5) = (2*y - 5)/(2*y + 5), answer
6*y^2 - 13*y - 5 = (3*y + 1)*(2*y - 5)
3*y^2 + 10*y - 8 = (y + 4)*(3*y - 2)
2*y^2 + 13*y + 20 = (y + 4)*(2*y + 5)
(3*y + 1)*(3*y - 2)/(3*y + 1)*(2*y - 5) = (3*y - 2)/(2*y - 5), answer
(y + 4)*(3*y - 2)/(y + 4)*(2*y + 5) = (3*y - 2)/(2*y + 5)
(3*y - 2)/(2*y - 5)/(3*y - 2)/(2*y + 5) = (2*y - 5)/(2*y + 5), answer
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have you even tried googling it?
http://en.wikipedia.org/wiki/Polynomial_…
http://en.wikipedia.org/wiki/Polynomial_…