Find the Holes in the Graph (x^2-5x+6)/(x^2-4x+3)

$\frac{x^{2} - 5 x + 6}{x^{2} - 4 x + 3}$

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$\frac{(x - 3) (x - 2)}{x^{2} - 4 x + 3}$

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 3$

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

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{x - 2}{x - 1}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $2$ from $3$ .

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

Subtract $1$ from $3$ .

$\frac{1}{2}$

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

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

Find the Holes in the Graph (x^2-5x+6)/(x^2-4x+3)

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