Find the Holes in the Graph (-3x^2+75)/(5x^2+35x+50)

$\frac{- 3 x^{2} + 75}{5 x^{2} + 35 x + 50}$

Factor $- 3 x^{2} + 75$ .

Factor $3$ out of $- 3 x^{2} + 75$ .

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

$\frac{3 (- x^{2}) + 75}{5 x^{2} + 35 x + 50}$

Factor $3$ out of $75$ .

$\frac{3 (- x^{2}) + 3 (25)}{5 x^{2} + 35 x + 50}$

Factor $3$ out of $3 (- x^{2}) + 3 (25)$ .

$\frac{3 (- x^{2} + 25)}{5 x^{2} + 35 x + 50}$

Rewrite $25$ as $5^{2}$ .

$\frac{3 (- x^{2} + 5^{2})}{5 x^{2} + 35 x + 50}$

Reorder $- x^{2}$ and $5^{2}$ .

$\frac{3 (5^{2} - x^{2})}{5 x^{2} + 35 x + 50}$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$\frac{3 ((5 + x) (5 - x))}{5 x^{2} + 35 x + 50}$

Remove unnecessary parentheses.

$\frac{3 (5 + x) (5 - x)}{5 x^{2} + 35 x + 50}$

Factor $5 x^{2} + 35 x + 50$ .

Factor $5$ out of $5 x^{2} + 35 x + 50$ .

Factor $5$ out of $5 x^{2}$ .

$\frac{3 (5 + x) (5 - x)}{5 (x^{2}) + 35 x + 50}$

Factor $5$ out of $35 x$ .

$\frac{3 (5 + x) (5 - x)}{5 (x^{2}) + 5 (7 x) + 50}$

Factor $5$ out of $50$ .

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

Factor $5$ out of $5 x^{2} + 5 (7 x)$ .

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

Factor $5$ out of $5 (x^{2} + 7 x) + 5 \cdot 10$ .

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

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.

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

Remove unnecessary parentheses.

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

Cancel the common factor of $5 + x$ and $x + 5$ .

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

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

Simplify.

Cancel the common factor of $5 + 5$ and $5$ .

Factor $5$ out of $3 (5 + 5)$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Add $1$ and $1$ .

$\frac{3 \cdot 2}{- 5 + 2}$

Add $- 5$ and $2$ .

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

Multiply $3$ by $2$ .

$\frac{6}{- 3}$

Divide $6$ by $- 3$ .

$- 2$

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

$(- 5, - 2)$

Find the Holes in the Graph (-3x^2+75)/(5x^2+35x+50)

Related Questions

Support LCMGCF.com by disabling your adblocker.