Find the least common denominators of the rational expressions: 2 x 2 + 5 x − 3 x 2 − 9 , 4 x − 12 x 2 − x − 6 , x 2 + x − 2 x 2 − 4 x + 3 \frac{2x^2+5x-3}{x^2-9},\frac{4x-12}{x^2-x-6},\frac{x^2+x-2}{x^2-4x+3}

This problem asked us to find the least common denominator of the rational expressions, and we're now moving on to part c. Now this problem looks a little bit complicated because I can see that we are now dealing with three rational expressions. But don't sweat it because we can solve this exactly like how we've solved the other problems when we're trying to find the least common denominator. So the same way we find the LCD for two rational expressions, we can use the same strategy for three, where I really just need to factor all the denominators that I see. I'm going to start with this first rational expression, two x squared plus five x minus three. In the denominator, notice we have x squared minus nine, which is the same thing as x squared minus three squared. Nine can be written as three squared. Now, of course, this will break down into a difference of squares. So this is the same thing as x plus three times x minus three. And next, I'll move on to the second rational expression where we have four x minus 12 in the numerator and x squared minus x minus six in the denominator. Well, I can see here that we have a quadratic polynomial with one as the leading coefficient. So I'm looking for a pair of numbers that multiplies to give me this negative six and adds to give me this negative one. Well, thinking about that, that's going to be two and negative three. Two times negative three is negative six. Two plus negative three is negative one. So we're going to get x plus two times x minus three. Now lastly, I'll move to this last rational expression where we have x squared plus x minus two in the numerator. And then in the denominator, we wanna factor this rational or we wanna factor this denominator. But I can see that we, once again, have a quadratic polynomial with one as the leading coefficient in front of x squared, meaning I can just think of a pair of numbers that multiplies to give me positive three and adds to give me negative four. Well, if I think about it, that's going to be negative one and negative three. Those multiply to give me positive three and add to give me negative four, giving me x minus one times x minus three. Now let's find the least common denominator. Well, I can see that we have an x plus three. I don't see an x plus three anywhere else in any of these denominators, so that x plus three is a unique prime factor. Now let's move on to this x minus three. I can see that x minus three is a common term that we have in all three of these denominators. So x minus three by itself will be a unique prime factor. There's no other x minus three's we need to write, we only need to write it once. And I can also see we have an x plus two. I don't see an x plus two anywhere else, so x plus two is a unique prime factor. Now, I also see that we have an x minus one. I don't see an x minus one anywhere else, so x minus one is a unique prime factor, and this would be the LCD and the solution to part c of this problem. Hope you found this video helpful.

Find the least common denominators of the rational expressions: 2 x 2 + 5 x − 3 x 2 − 9 , 4 x − 12 x 2 − x − 6 , x 2 + x − 2 x 2 − 4 x + 3 \frac{2x^2+5x-3}{x^2-9},\frac{4x-12}{x^2-x-6},\frac{x^2+x-2}{x^2-4x+3}

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

( x − 3 ) ( x − 3 ) 2 ( x − 2 ) \left(x-3\right)\left(x-3\right)^2\left(x-2\right) ( x − 3 ) ( x − 3 ) 2 ( x − 2 )

8. Rational Expressions and Functions

Least Common Denominators

Intermediate Algebra

8. Rational Expressions and Functions

Least Common Denominators

Struggling with Intermediate Algebra?

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Multiple Choice

Find the least common denominators of the rational expressions:
$\frac{2 x^{2} + 5 x - 3}{x^{2} - 9}, \frac{4 x - 12}{x^{2} - x - 6}, \frac{x^{2} + x - 2}{x^{2} - 4 x + 3}$

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

$\left(x+3\right)\left(x-3\right)^2\left(x+2\right)$

$\left(x-3\right)\left(x-3\right)^2\left(x-2\right)$

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

Verified step by step guidance

Start by factoring each denominator completely to identify their prime factors. For the first denominator $x^2 - 9$, recognize it as a difference of squares and factor it as $\left(x+3\right)\left(x-3\right)$.

Next, factor the second denominator $x^2 - x - 6$. Find two numbers that multiply to $-6$ and add to $-1$. This factors as $\left(x-3\right)\left(x+2\right)$.

Then, factor the third denominator $x^2 - 4x + 3$. Find two numbers that multiply to \$3$ and add to $-4$. This factors as $\left(x-3\right)\left(x-1\right)$.

To find the least common denominator (LCD), take each distinct factor from all denominators, using the highest power of each factor that appears. Here, the factors are $\left(x+3\right)$, $\left(x-3\right)$, $\left(x+2\right)$, and $\left(x-1\right)$. Note that $\left(x-3\right)$ appears in all denominators but only to the first power, so include it once.

Multiply these factors together to write the LCD as $\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x-1\right)$. This expression represents the least common denominator for the given rational expressions.

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Struggling with Intermediate Algebra?

Find the least common denominators of the rational expressions:2x2+5x−3x2−9,4x−12x2−x−6,x2+x−2x2−4x+3\frac{2x^2+5x-3}{x^2-9},\frac{4x-12}{x^2-x-6},\frac{x^2+x-2}{x^2-4x+3}

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Find the least common denominators of the rational expressions:
$\frac{2 x^{2} + 5 x - 3}{x^{2} - 9}, \frac{4 x - 12}{x^{2} - x - 6}, \frac{x^{2} + x - 2}{x^{2} - 4 x + 3}$