College Algebra (10th Edition)

$\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$
Step 1: Factor each denominator For $x^{2}+bx+c$, we search for factors of c whose sum is b: $x^{2}-7x+6 = \quad$ ... -6 and -1... $=(x-6)(x-1)$ $x^{2}-2x-24 = \quad$ ... $-6$ and $+4$... $=(x-6)(x+4)$ Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. LCM = $(x-6)(x-1)(x+4)$ Step 3: Write each rational expression using the LCM as the denominator. $\displaystyle \frac{x}{ (x-6)(x-1)}-\frac{x}{ (x-6)(x+4)}$= $= \displaystyle \frac{x(x+4)}{ (x-6)(x-1)(x+4)}-\frac{x(x-1)}{ (x-6)(x+4)(x-1)}$= $=\displaystyle \frac{x^{2}+4x-(x^{2}-x)}{ (x-6)(x-1)(x+4)}$ $=\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$