Find the Holes in the Graph ((2x-1)(x+4)^2(x-1)(3x+2))/(3x(x-4)(2x-1)^3)

$\frac{(2 x - 1) {(x + 4)}^{2} (x - 1) (3 x + 2)}{3 x (x - 4) {(2 x - 1)}^{3}}$

Step 1

Cancel the common factor of $2 x - 1$ and ${(2 x - 1)}^{3}$ .

Factor $2 x - 1$ out of $(2 x - 1) {(x + 4)}^{2} (x - 1) (3 x + 2)$ .

$\frac{(2 x - 1) (({(x + 4)}^{2} (x - 1)) (3 x + 2))}{3 x (x - 4) {(2 x - 1)}^{3}}$

Cancel the common factors.

Factor $2 x - 1$ out of $3 x (x - 4) {(2 x - 1)}^{3}$ .

$\frac{(2 x - 1) (({(x + 4)}^{2} (x - 1)) (3 x + 2))}{(2 x - 1) (3 x (x - 4) {(2 x - 1)}^{2})}$

Cancel the common factor.

$\frac{(2 x - 1) (({(x + 4)}^{2} (x - 1)) (3 x + 2))}{(2 x - 1) (3 x (x - 4) {(2 x - 1)}^{2})}$

Rewrite the expression.

$\frac{({(x + 4)}^{2} (x - 1)) (3 x + 2)}{3 x (x - 4) {(2 x - 1)}^{2}}$

Step 2

Multiply ${(x + 4)}^{2} (x - 1)$ by $3 x + 2$ .

$\frac{{(x + 4)}^{2} (x - 1) (3 x + 2)}{3 x (x - 4) {(2 x - 1)}^{2}}$

Step 3

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes

Step 4