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

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

Step 1

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

Step 2

Cancel the common factor of $x + 2$ and ${(x + 2)}^{4}$ .

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Cancel the common factor of $2 x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

Multiply $4$ by $3 x + 4$ .

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

Step 6

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 1, 2 x - 3$

Step 7

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{4 (3 x + 4)}{(3 (x - 2) (x - 1)) {(x + 2)}^{3}}$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Substitute $- 1$ for $x$ in $\frac{4 (3 x + 4)}{(3 (x - 2) (x - 1)) {(x + 2)}^{3}}$ and simplify.

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Simplify the numerator.

Multiply $3$ by $- 1$ .

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

Add $- 3$ and $4$ .

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

Simplify the denominator.

Subtract $2$ from $- 1$ .

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

Multiply $3$ by $- 3$ .

$\frac{4 \cdot 1}{- 9 (- 1 - 1) {(- 1 + 2)}^{3}}$

Subtract $1$ from $- 1$ .

$\frac{4 \cdot 1}{- 9 \cdot - 2 {(- 1 + 2)}^{3}}$

Multiply $- 9$ by $- 2$ .

$\frac{4 \cdot 1}{18 {(- 1 + 2)}^{3}}$

Add $- 1$ and $2$ .

$\frac{4 \cdot 1}{18 \cdot 1^{3}}$

One to any power is one.

$\frac{4 \cdot 1}{18 \cdot 1}$

Multiply $4$ by $1$ .

$\frac{4}{18 \cdot 1}$

Multiply $18$ by $1$ .

$\frac{4}{18}$

Cancel the common factor of $4$ and $18$ .

Factor $2$ out of $4$ .

$\frac{2 (2)}{18}$

Cancel the common factors.

Factor $2$ out of $18$ .

$\frac{2 \cdot 2}{2 \cdot 9}$

Cancel the common factor.

$\frac{2 \cdot 2}{2 \cdot 9}$

Rewrite the expression.

$\frac{2}{9}$

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Substitute $\frac{3}{2}$ for $x$ in $\frac{4 (3 x + 4)}{(3 (x - 2) (x - 1)) {(x + 2)}^{3}}$ and simplify.

Substitute $\frac{3}{2}$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Simplify the numerator.

Multiply $3 (\frac{3}{2})$ .

Combine $3$ and $\frac{3}{2}$ .

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

Multiply $3$ by $3$ .

$\frac{4 (\frac{9}{2} + 4)}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 (\frac{9}{2} + 4 \cdot \frac{2}{2})}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine $4$ and $\frac{2}{2}$ .

$\frac{4 (\frac{9}{2} + \frac{4 \cdot 2}{2})}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine the numerators over the common denominator.

$\frac{4 \frac{9 + 4 \cdot 2}{2}}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Simplify the numerator.

Multiply $4$ by $2$ .

$\frac{4 \frac{9 + 8}{2}}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Add $9$ and $8$ .

$\frac{4 (\frac{17}{2})}{3 (\frac{3}{2} - 2) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Simplify the denominator.

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{3 (\frac{3}{2} - 2 \cdot \frac{2}{2}) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine $- 2$ and $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{3 (\frac{3}{2} + \frac{- 2 \cdot 2}{2}) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine the numerators over the common denominator.

$\frac{4 (\frac{17}{2})}{3 \frac{3 - 2 \cdot 2}{2} (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Simplify the numerator.

Multiply $- 2$ by $2$ .

$\frac{4 (\frac{17}{2})}{3 \frac{3 - 4}{2} (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Subtract $4$ from $3$ .

$\frac{4 (\frac{17}{2})}{3 (\frac{- 1}{2}) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Move the negative in front of the fraction.

$\frac{4 (\frac{17}{2})}{3 \cdot - 1 \frac{1}{2} (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine exponents.

Multiply $3$ by $- 1$ .

$\frac{4 (\frac{17}{2})}{- 3 (\frac{1}{2}) (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine $- 3$ and $\frac{1}{2}$ .

$\frac{4 (\frac{17}{2})}{\frac{- 3}{2} (\frac{3}{2} - 1) {(\frac{3}{2} + 2)}^{3}}$

Combine ${(\frac{3}{2} + 2)}^{3}$ and $\frac{- 3}{2}$ .

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{3}{2} + 2)}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Simplify the numerator.

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{3}{2} + 2 \cdot \frac{2}{2})}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Combine $2$ and $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{3}{2} + \frac{2 \cdot 2}{2})}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Combine the numerators over the common denominator.

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{3 + 2 \cdot 2}{2})}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{3 + 4}{2})}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Add $3$ and $4$ .

$\frac{4 (\frac{17}{2})}{\frac{{(\frac{7}{2})}^{3} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Apply the product rule to $\frac{7}{2}$ .

$\frac{4 (\frac{17}{2})}{\frac{\frac{7^{3}}{2^{3}} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Raise $7$ to the power of $3$ .

$\frac{4 (\frac{17}{2})}{\frac{\frac{343}{2^{3}} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Raise $2$ to the power of $3$ .

$\frac{4 (\frac{17}{2})}{\frac{\frac{343}{8} \cdot - 3}{2} (\frac{3}{2} - 1)}$

Combine $\frac{343}{8}$ and $- 3$ .

$\frac{4 (\frac{17}{2})}{\frac{\frac{343 \cdot - 3}{8}}{2} (\frac{3}{2} - 1)}$

Multiply $343$ by $- 3$ .

$\frac{4 (\frac{17}{2})}{\frac{\frac{- 1029}{8}}{2} (\frac{3}{2} - 1)}$

Move the negative in front of the fraction.

$\frac{4 (\frac{17}{2})}{\frac{- \frac{1029}{8}}{2} (\frac{3}{2} - 1)}$

Multiply the numerator by the reciprocal of the denominator.

$\frac{4 (\frac{17}{2})}{- \frac{1029}{8} \cdot \frac{1}{2} (\frac{3}{2} - 1)}$

Multiply $- \frac{1029}{8} \cdot \frac{1}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{1029}{8}$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{2 \cdot 8} (\frac{3}{2} - 1)}$

Multiply $2$ by $8$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} (\frac{3}{2} - 1)}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} (\frac{3}{2} - 1 \cdot \frac{2}{2})}$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} (\frac{3}{2} + \frac{- 1 \cdot 2}{2})}$

Combine the numerators over the common denominator.

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} \cdot \frac{3 - 1 \cdot 2}{2}}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} \cdot \frac{3 - 2}{2}}$

Subtract $2$ from $3$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{16} \cdot \frac{1}{2}}$

Combine exponents.

Multiply $\frac{1}{2}$ by $\frac{1029}{16}$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{2 \cdot 16}}$

Multiply $2$ by $16$ .

$\frac{4 (\frac{17}{2})}{- \frac{1029}{32}}$

Combine $4$ and $\frac{17}{2}$ .

$\frac{\frac{4 \cdot 17}{2}}{- \frac{1029}{32}}$

Multiply $4$ by $17$ .

$\frac{\frac{68}{2}}{- \frac{1029}{32}}$

Divide $68$ by $2$ .

$\frac{34}{- \frac{1029}{32}}$

Multiply the numerator by the reciprocal of the denominator.

$34 (- \frac{32}{1029})$

Multiply $34 (- \frac{32}{1029})$ .

Multiply $- 1$ by $34$ .

$- 34 (\frac{32}{1029})$

Combine $- 34$ and $\frac{32}{1029}$ .

$\frac{- 34 \cdot 32}{1029}$

Multiply $- 34$ by $32$ .

$\frac{- 1088}{1029}$

Move the negative in front of the fraction.

$- \frac{1088}{1029}$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(- 1, \frac{2}{9}), (\frac{3}{2}, - \frac{1088}{1029})$

Step 8

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

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