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

Factor ${\displaystyle 3}$ out of ${\displaystyle -3x}$.

Factor ${\displaystyle 3}$ out of ${\displaystyle 3}$.

Factor ${\displaystyle 3}$ out of ${\displaystyle 3(-x)+3\left(1\right)}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 5}$ and whose sum is ${\displaystyle -6}$.

Write the factored form using these integers.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle 1}$ as ${\displaystyle -1(-1)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1(-1)}$.

Rewrite ${\displaystyle -(x-1)}$ as ${\displaystyle -1(x-1)}$.

Cancel the common factor.

Rewrite the expression.

Multiply ${\displaystyle 3}$ by ${\displaystyle -1}$.

Multiply ${\displaystyle 3}$ by ${\displaystyle -1}$.

Set ${\displaystyle x-1}$ equal to ${\displaystyle 0}$.

Add ${\displaystyle 1}$ to both sides of the equation.

Substitute ${\displaystyle 1}$ for ${\displaystyle x}$ in ${\displaystyle -\frac{3}{x-5}}$ and simplify.

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