Find the Holes in the Graph (2x^2-5x+2)/(x^2-4)

$\frac{2 x^{2} - 5 x + 2}{x^{2} - 4}$

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 $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$\frac{2 x^{2} - 5 (x) + 2}{x^{2} - 4}$

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Apply the distributive property.

$\frac{2 x^{2} - 1 x - 4 x + 2}{x^{2} - 4}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$\frac{x (2 x - 1) - 2 (2 x - 1)}{x^{2} - 4}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{(2 x - 1) (x - 2)}{x^{2} - 4}$

Factor $x^{2} - 4$ .

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

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

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 $b = 2$ .

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 x - 1}{x + 2}$

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 $\frac{2 x - 1}{x + 2}$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$\frac{2 \cdot 2 - 1}{2 + 2}$

Simplify.

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{4 - 1}{2 + 2}$

Subtract $1$ from $4$ .

$\frac{3}{2 + 2}$

Add $2$ and $2$ .

$\frac{3}{4}$

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

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

Find the Holes in the Graph (2x^2-5x+2)/(x^2-4)

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