Find the Holes in the Graph (x^2-4)/(2x^2-11x+14)

$\frac{x^{2} - 4}{2 x^{2} - 11 x + 14}$

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

$\frac{x^{2} - 2^{2}}{2 x^{2} - 11 x + 14}$

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 $b = 2$ .

$\frac{(x + 2) (x - 2)}{2 x^{2} - 11 x + 14}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 14 = 28$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$\frac{(x + 2) (x - 2)}{2 x^{2} - 11 (x) + 14}$

Rewrite $- 11$ as $- 4$ plus $- 7$

$\frac{(x + 2) (x - 2)}{2 x^{2} + (- 4 - 7) x + 14}$

Apply the distributive property.

$\frac{(x + 2) (x - 2)}{2 x^{2} - 4 x - 7 x + 14}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x + 2) (x - 2)}{(2 x^{2} - 4 x) - 7 x + 14}$

Factor out the greatest common factor (GCF) from each group.

$\frac{(x + 2) (x - 2)}{2 x (x - 2) - 7 (x - 2)}$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$\frac{(x + 2) (x - 2)}{(x - 2) (2 x - 7)}$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{(x + 2) (x - 2)}{(x - 2) (2 x - 7)}$

Rewrite the expression.

$\frac{x + 2}{2 x - 7}$

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

$x - 2$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 2}{2 x - 7}$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $\frac{x + 2}{2 x - 7}$ and simplify.

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

$\frac{2 + 2}{2 \cdot 2 - 7}$

Simplify.

Add $2$ and $2$ .

$\frac{4}{2 \cdot 2 - 7}$

Simplify the denominator.

Multiply $2$ by $2$ .

$\frac{4}{4 - 7}$

Subtract $7$ from $4$ .

$\frac{4}{- 3}$

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 $0$ .

$(2, - \frac{4}{3})$

Find the Holes in the Graph (x^2-4)/(2x^2-11x+14)

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