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

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

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

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

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

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

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

Factor $2 x$ out of $- 2 x$ .

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

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

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 4}{2 x}$

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 $\frac{x - 4}{2 x}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $\frac{x - 4}{2 x}$ and simplify.

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

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

Simplify.

Subtract $4$ from $1$ .

$\frac{- 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$\frac{- 3}{2}$

Move the negative in front of the fraction.

$- \frac{3}{2}$

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

$(1, - \frac{3}{2})$

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

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