Question

What is the least common denominator of the rational expression: `5/x^2 - 3/(6x^2 + 12x)`?

Answer 1

The first fraction is set, yet the second one needs simplifying- which I missed pre-edit. `3/(6x^2+12x)= 3/(6x(x+2))=1/(2x(x+2)`. Then we compare leftover denominators to find the LCD of `x^2` and `2x(x+2)` getting `2x^2(x+2)=2x^3+4x^2`. What the other guys have

Answer 2

`2x^3+4x^2`

Explanation:

The second term is not in minimal terms: there is a factor `3` that can be taken out:

`frac{3}{6x^2+12x}=(frac{3}{3})(frac{1}{2x^3+4x})`

You now can use the formula

`lcm(a,b)=frac{ab}{GCD(a,b)}`

Since `GCD(x^2,(2x^2+4x))=x`, we have that

`lcm(x^2, (2x^2+4x))=frac{x^2(2x^2+4x)}{x}=2x^3+4x^2`

Hence your difference becomes

`frac{5(2x+4)}{2x^3+4x^2}-frac{x}{2x^3+4x^2}=frac{9x+20}{2x^3+4x^2}`

Answer 3

`2x^3-4x^2`

Explanation:

To adjust the fractions to common denominators so the terms can be combined, you would want to multiply each fraction by the number 1 in the form of the other fraction's denominator. I notice that 6x^2+12x can be factored to 6x(x+2) and x^2 is x*x, So, and x is already in common.

The left fraction, we would multiply the top and bottom by 6x+12, and the right fraction by x.

`5(6x+12)/(x^2(6x+12))-3x/(x*x(6x+12))=(27x+60)/(6x^2(x+2))=(9x+20)/(2x^2(x+2))`