Find the Holes in the Graph (5x^2-10x+5)/(2x^2-x-1)

$\frac{5 x^{2} - 10 x + 5}{2 x^{2} - x - 1}$

Factor $5 x^{2} - 10 x + 5$ .

Factor $5$ out of $5 x^{2} - 10 x + 5$ .

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

$\frac{5 (x^{2}) - 10 x + 5}{2 x^{2} - x - 1}$

Factor $5$ out of $- 10 x$ .

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

Factor $5$ out of $5$ .

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

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

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

Factor $5$ out of $5 (x^{2} - 2 x) + 5 (1)$ .

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $- 1$ as $1$ plus $- 2$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Cancel the common factor of ${(x - 1)}^{2}$ and $x - 1$ .

Factor $x - 1$ out of $5 {(x - 1)}^{2}$ .

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

Cancel the common factors.

Factor $x - 1$ out of $(2 x + 1) (x - 1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 1$

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

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $1$ from $1$ .

$\frac{5 \cdot 0}{2 \cdot 1 + 1}$

Simplify the denominator.

Multiply $2$ by $1$ .

$\frac{5 \cdot 0}{2 + 1}$

Add $2$ and $1$ .

$\frac{5 \cdot 0}{3}$

Multiply $5$ by $0$ .

$\frac{0}{3}$

Divide $0$ by $3$ .

$0$

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

$(1, 0)$

Find the Holes in the Graph (5x^2-10x+5)/(2x^2-x-1)

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