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

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

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

Simplify.

Simplify the numerator.

Subtract $3$ from $1$ .

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

Add $1$ and $1$ .

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

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

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

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

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

Simplify the denominator.

Subtract $2$ from $1$ .

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

Add $1$ and $2$ .

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

Multiply $- 1$ by $3$ .

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

Add $1$ and $3$ .

$\frac{4 \cdot 4}{- 3 \cdot 4}$

Multiply $4$ by $4$ .

$\frac{16}{- 3 \cdot 4}$

Multiply $- 3$ by $4$ .

$\frac{16}{- 12}$

Cancel the common factor of $16$ and $- 12$ .

Factor $4$ out of $16$ .

$\frac{4 (4)}{- 12}$

Cancel the common factors.

Factor $4$ out of $- 12$ .

$\frac{4 \cdot 4}{4 \cdot - 3}$

Cancel the common factor.

$\frac{4 \cdot 4}{4 \cdot - 3}$

Rewrite the expression.

$\frac{4}{- 3}$

Move the negative in front of the fraction.

$- \frac{4}{3}$

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

$(1, - \frac{4}{3})$

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

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