Find the least common denominators of the rational expressions:2y2+3y−5y2−4\frac{2y^2+3y-5}{y^2-4}, y−2y2+y−6\frac{y-2}{y^2+y-6}

( y + 2 ) ( y − 2 ) ( y + 3 ) \left(y+2\right)\left(y-2\right)\left(y+3\right)

Find the least common denominators of the rational expressions: 2 y 2 + 3 y − 5 y 2 − 4 \frac{2y^2+3y-5}{y^2-4} , y − 2 y 2 + y − 6 \frac{y-2}{y^2+y-6}

This problem wants us to find the least common denominators of the rational expressions. Now we're starting with part a of this problem, where we have these two rational expressions, and we want to find the least common denominator. Now the least common denominator is gonna be the product of all the unique terms that we can find. Now the best way to do this is to strictly factor these denominators here to see what terms or factors are unique. Now starting with this first fraction where we have two y squared plus three y minus five on top, notice that we have y squared minus four. Y squared minus four is the same thing as y squared minus two squared. And this is the difference between two squares. So y squared minus two squared can be factored into y plus two times y minus two. And that's what happens when we fully factor the denominator for this first fraction that we see. Now for the second fraction where we have, or I should say rational expression, where we have y minus two on top, in the denominator we're going to have y squared plus y minus six. What I can do is think about a pair of numbers that multiply to give me negative six and add to give me this one coefficient in front of the y. Because that's what I'm looking for since I can see we have a one coefficient in front of the y squared, I can just factor this like a typical quadratic polynomial. And And if I think about that pair of numbers, well, that's going to be negative two and three. Because negative two times three gives me negative six. Negative two plus three gives me positive one in front of this y. So that means this is going to be factored into x minus or y minus two, I should say, y minus two times y plus three. So we now have the denominators fully factored, and I need to look for unique terms or unique factors in the denominator. Now you can see we have a y plus two right here, and I don't see a y plus two anywhere else. So y plus two is unique. Now you can see we have a y minus two, but we also have a y minus two there. So I'm only gonna write that y minus two once. I don't need to write it twice. And then I can see we have a y plus three. That's the only y plus three that we have. So this means that y plus three will be this last term, and multiplying those all across will give me the LCD as a solution to part a of this problem.

Find the least common denominators of the rational expressions: 2 y 2 + 3 y − 5 y 2 − 4 \frac{2y^2+3y-5}{y^2-4} , y − 2 y 2 + y − 6 \frac{y-2}{y^2+y-6}

( y + 2 ) ( y − 2 ) ( y + 3 ) \left(y+2\right)\left(y-2\right)\left(y+3\right)

( y − 2 ) ( y − 2 ) ( y + 3 ) \left(y-2\right)\left(y-2\right)\left(y+3\right)

8. Rational Expressions and Equations

Least Common Denominators

Beginning Algebra

8. Rational Expressions and Equations

Least Common Denominators

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Multiple Choice

Find the least common denominators of the rational expressions:
$\frac{2 y^{2} + 3 y - 5}{y^{2} - 4}$ , $\frac{y - 2}{y^{2} + y - 6}$

$(y - 2) (y - 2) (y + 3)$

$y squared minus 4$

$y^{2} + y - 6$

$(y + 2) (y - 2) (y + 3)$

Verified step by step guidance

Identify the denominators of the given rational expressions: the first denominator is $y^2 - 4$ and the second denominator is $y^2 + y - 6$.

Factor each denominator completely. For $y^2 - 4$, recognize it as a difference of squares and factor it as $(y - 2)(y + 2)$.

For $y^2 + y - 6$, find two numbers that multiply to $-6$ and add to \$1$. This factors as $(y + 3)(y - 2)$.

To find the least common denominator (LCD), take each distinct factor from both denominators, using the highest power of each factor that appears. The factors are $(y - 2)$, $(y + 2)$, and $(y + 3)$.

Multiply these factors together to express the LCD as $(y + 2)(y - 2)(y + 3)$.

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