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

Feb 16, 2018

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 $2 x \left(x + 2\right)$ getting $2 {x}^{2} \left(x + 2\right) = 2 {x}^{3} + 4 {x}^{2}$. What the other guys have

Feb 16, 2018

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

#### Explanation:

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

$\frac{3}{6 {x}^{2} + 12 x} = \left(\frac{3}{3}\right) \left(\frac{1}{2 {x}^{3} + 4 x}\right)$

You now can use the formula

$\lcm \left(a , b\right) = \frac{a b}{G C D \left(a , b\right)}$

Since $G C D \left({x}^{2} , \left(2 {x}^{2} + 4 x\right)\right) = x$, we have that

$\lcm \left({x}^{2} , \left(2 {x}^{2} + 4 x\right)\right) = \frac{{x}^{2} \left(2 {x}^{2} + 4 x\right)}{x} = 2 {x}^{3} + 4 {x}^{2}$

$\frac{5 \left(2 x + 4\right)}{2 {x}^{3} + 4 {x}^{2}} - \frac{x}{2 {x}^{3} + 4 {x}^{2}} = \frac{9 x + 20}{2 {x}^{3} + 4 {x}^{2}}$

Feb 16, 2018

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

#### 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.

$5 \frac{6 x + 12}{{x}^{2} \left(6 x + 12\right)} - 3 \frac{x}{x \cdot x \left(6 x + 12\right)} = \frac{27 x + 60}{6 {x}^{2} \left(x + 2\right)} = \frac{9 x + 20}{2 {x}^{2} \left(x + 2\right)}$