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

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

Step 1

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 2

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

$x + 2$

Step 3

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

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Simplify the numerator.

Add $- 2$ and $1$ .

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

Raise $- 1$ to the power of $3$ .

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

Simplify the denominator.

Subtract $3$ from $- 2$ .

$\frac{- 1}{{(- 5)}^{2}}$

Raise $- 5$ to the power of $2$ .

$\frac{- 1}{25}$

Move the negative in front of the fraction.

$- \frac{1}{25}$

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

$(- 2, - \frac{1}{25})$

Step 4

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

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