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}

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

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: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{$$x^2+x-2$$}{$$x^2-4x+3$$}

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

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

(x+3)(x−3)(x+2)(x+1)\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x+1\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:
$\frac{2 x^{2} + 5 x - 3}{x^{2} - 9}, \frac{4 x - 12}{x^{2} - x - 6}, \frac{x^{2} + x - 2}{x^{2} - 4 x + 3}$

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

$(x + 3) {(x - 3)}^{2} (x + 2)$

$(x - 3) {(x - 3)}^{2} (x - 2)$

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

Verified step by step guidance

Identify the denominators of the given rational expressions: $x^2 - 9$, $x^2 - x - 6$, and $x^2 - 4x + 3$.

Factor each denominator completely using factoring techniques such as difference of squares and factoring trinomials: - $x^2 - 9$ factors as $(x+3)(x-3)$, - $x^2 - x - 6$ factors as $(x-3)(x+2)$, - $x^2 - 4x + 3$ factors as $(x-3)(x-1)$.

List all the unique factors from the factorizations: $(x+3)$, $(x-3)$, $(x+2)$, and $(x-1)$.

Determine the least common denominator (LCD) by taking each unique factor at its highest power found in any denominator. Since $(x-3)$ appears only once in each denominator, include it once.

Write the LCD as the product of these factors: $(x+3)(x-3)(x+2)(x-1)$.

5:30m

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Struggling with Beginning 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:
$\frac{2 x^{2} + 5 x - 3}{x^{2} - 9}, \frac{4 x - 12}{x^{2} - x - 6}, \frac{x^{2} + x - 2}{x^{2} - 4 x + 3}$