Find the Holes in the Graph f(x)=(x^2+4x-5)/(3x^2+21x+30)

$f (x) = \frac{x^{2} + 4 x - 5}{3 x^{2} + 21 x + 30}$

Factor $x^{2} + 4 x - 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$f (x) = \frac{(x - 1) (x + 5)}{3 x^{2} + 21 x + 30}$

Factor $3 x^{2} + 21 x + 30$ .

Factor $3$ out of $3 x^{2} + 21 x + 30$ .

Factor $3$ out of $3 x^{2}$ .

$f (x) = \frac{(x - 1) (x + 5)}{3 (x^{2}) + 21 x + 30}$

Factor $3$ out of $21 x$ .

$f (x) = \frac{(x - 1) (x + 5)}{3 (x^{2}) + 3 (7 x) + 30}$

Factor $3$ out of $30$ .

$f (x) = \frac{(x - 1) (x + 5)}{3 x^{2} + 3 (7 x) + 3 \cdot 10}$

Factor $3$ out of $3 x^{2} + 3 (7 x)$ .

$f (x) = \frac{(x - 1) (x + 5)}{3 (x^{2} + 7 x) + 3 \cdot 10}$

Factor $3$ out of $3 (x^{2} + 7 x) + 3 \cdot 10$ .

$f (x) = \frac{(x - 1) (x + 5)}{3 (x^{2} + 7 x + 10)}$

Factor.

Factor $x^{2} + 7 x + 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$f (x) = \frac{(x - 1) (x + 5)}{3 ((x + 2) (x + 5))}$

Remove unnecessary parentheses.

$f (x) = \frac{(x - 1) (x + 5)}{3 (x + 2) (x + 5)}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f (x) = \frac{(x - 1) (x + 5)}{3 (x + 2) (x + 5)}$

Rewrite the expression.

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

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

$x + 5$

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

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Substitute $- 5$ for $x$ in $\frac{x - 1}{3 (x + 2)}$ and simplify.

Substitute $- 5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 5 - 1}{3 (- 5 + 2)}$

Simplify.

Subtract $1$ from $- 5$ .

$\frac{- 6}{3 (- 5 + 2)}$

Add $- 5$ and $2$ .

$\frac{- 6}{3 \cdot - 3}$

Multiply $3$ by $- 3$ .

$\frac{- 6}{- 9}$

Cancel the common factor of $- 6$ and $- 9$ .

Factor $- 3$ out of $- 6$ .

$\frac{- 3 (2)}{- 9}$

Cancel the common factors.

Factor $- 3$ out of $- 9$ .

$\frac{- 3 \cdot 2}{- 3 \cdot 3}$

Cancel the common factor.

$\frac{- 3 \cdot 2}{- 3 \cdot 3}$

Rewrite the expression.

$\frac{2}{3}$

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

$(- 5, \frac{2}{3})$

Find the Holes in the Graph f(x)=(x^2+4x-5)/(3x^2+21x+30)

Related Questions

Support LCMGCF.com by disabling your adblocker.