Find the least common denominators of the rational expressions:3x2−12x2−x−6\frac{3x^2-12}{x^2-x-6}, 5x+10x2−4x+3\frac{5x+10}{x^2-4x+3}

( x + 2 ) ( x − 3 ) ( x − 1 ) \left(x+2\right)\left(x-3\right)\left(x-1\right)

Find the least common denominators of the rational expressions: 3 x 2 − 12 x 2 − x − 6 \frac{3x^2-12}{x^2-x-6} , 5 x + 10 x 2 − 4 x + 3 \frac{5x+10}{x^2-4x+3}

This problem asks us to find the least common denominators of the rational expressions, and we're now moving on to part b of this problem. Now to find the least common denominator, this is going to be a collection of all the unique prime factors that we can find multiplied all across. So the way that we can do this is by focusing on the denominators and factoring them completely. Now let's start with this denominator we have on the left side, where the numerator is three x squared minus 12. Notice this is a quadratic polynomial with one as the leading coefficient in front of x squared, so what I need to do is find a pair of numbers that multiply to give me negative six, and add to give me this negative one coefficient in front of the x. Now thinking about that pair of numbers, well, that's going to be two and negative three. Two times negative three gives me negative six. Two plus negative three gives me negative one. So what that's gonna give me is x plus two times x minus three. Now moving on to this next fraction that I see here, this next rational expression, we're gonna have five x plus 10 in the numerator, and then I need to factor the denominator. Notice that we have x squared minus 4x plus three. We can still see there's a one coefficient in front of the x squared, so we can just think about a pair of numbers that multiplies to give me three, but then adds to give me negative four. What pair of numbers would that be? Well, Well, if I think about it, that's just going to be negative one and negative three. These multiply to give me positive three, and they add to give me negative four. So this is going to break down into x minus one times x minus three. Now let's see what our least common denominator is going to be. Well, I can see we have an x plus two, but I don't see an x plus two anywhere else, nowhere in the denominators. So x plus two is a unique prime factor. And I see we have an x minus three, and I do see another x minus three that shows up there. We only need to write this once since we're looking for unique prime factors. And then I can see we have this x minus one, and this is the only x minus one that we have anywhere in these denominators, so we have x minus one. And this would be the LCD and the solution to part b of this problem.

Find the least common denominators of the rational expressions: 3 x 2 − 12 x 2 − x − 6 \frac{3x^2-12}{x^2-x-6} , 5 x + 10 x 2 − 4 x + 3 \frac{5x+10}{x^2-4x+3}

( x + 2 ) ( x − 3 ) ( x − 1 ) \left(x+2\right)\left(x-3\right)\left(x-1\right)

( x + 2 ) ( x − 3 ) ( x + 1 ) \left(x+2\right)\left(x-3\right)\left(x+1\right)

( x − 1 ) ( x − 3 ) \left(x-1\right)\left(x-3\right)

( x − 2 ) ( x + 3 ) \left(x-2\right)\left(x+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:
$\frac{3 x^{2} - 12}{x^{2} - x - 6}$ , $\frac{5 x + 10}{x^{2} - 4 x + 3}$

$(x + 2) (x - 3) (x + 1)$

$(x - 1) (x - 3)$

$(x + 2) (x - 3) (x - 1)$

$(x - 2) (x + 3)$

Verified step by step guidance

Start by factoring the denominators of both rational expressions to find their prime factors. For the first denominator $x^2 - x - 6$, look for two numbers that multiply to $-6$ and add to $-1$.

Factor the first denominator as $\left(x - 3\right)\left(x + 2\right)$ based on the numbers found.

Next, factor the second denominator $x^2 - 4x + 3$ by finding two numbers that multiply to \$3$ and add to $-4$.

Factor the second denominator as $\left(x - 3\right)\left(x - 1\right)$.

The least common denominator (LCD) is the product of all unique factors from both denominators. Combine $\left(x + 2\right)$, $\left(x - 3\right)$, and $\left(x - 1\right)$ to get the LCD: $\left(x + 2\right)\left(x - 3\right)\left(x - 1\right)$.

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