Find the Holes in the Graph (x^2-3x-10)/(2x^2+3x-2)

$\frac{x^{2} - 3 x - 10}{2 x^{2} + 3 x - 2}$

Step 1

Factor $x^{2} - 3 x - 10$ 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 $- 3$ .

$- 5, 2$

Write the factored form using these integers.

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

Step 2

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 $b = 3$ .

Factor $3$ out of $3 x$ .

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

Rewrite $3$ as $- 1$ plus $4$

$\frac{(x - 5) (x + 2)}{2 x^{2} + (- 1 + 4) x - 2}$

Apply the distributive property.

$\frac{(x - 5) (x + 2)}{2 x^{2} - 1 x + 4 x - 2}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 5) (x + 2)}{(2 x^{2} - 1 x) + 4 x - 2}$

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

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

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

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 5}{2 x - 1}$

Step 4

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

$x + 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 5}{2 x - 1}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$\frac{- 2 - 5}{2 \cdot - 2 - 1}$

Simplify.

Subtract $5$ from $- 2$ .

$\frac{- 7}{2 \cdot - 2 - 1}$

Simplify the denominator.

Multiply $2$ by $- 2$ .

$\frac{- 7}{- 4 - 1}$

Subtract $1$ from $- 4$ .

$\frac{- 7}{- 5}$

Dividing two negative values results in a positive value.

$\frac{7}{5}$

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

$(- 2, \frac{7}{5})$

Step 6

Find the Holes in the Graph (x^2-3x-10)/(2x^2+3x-2)

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