Find the Holes in the Graph f(x)=(x-5)/(3x^2-16x+5)

$f (x) = \frac{x - 5}{3 x^{2} - 16 x + 5}$

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 $a \cdot c = 3 \cdot 5 = 15$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 x$ .

$f (x) = \frac{x - 5}{3 x^{2} - 16 x + 5}$

Rewrite $- 16$ as $- 1$ plus $- 15$

$f (x) = \frac{x - 5}{3 x^{2} + (- 1 - 15) x + 5}$

Apply the distributive property.

$f (x) = \frac{x - 5}{3 x^{2} - 1 x - 15 x + 5}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{x - 5}{(3 x^{2} - 1 x) - 15 x + 5}$

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

$f (x) = \frac{x - 5}{x (3 x - 1) - 5 (3 x - 1)}$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$f (x) = \frac{x - 5}{(3 x - 1) (x - 5)}$

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$f (x) = \frac{x - 5}{(3 x - 1) (x - 5)}$

Rewrite the expression.

$f (x) = \frac{1}{3 x - 1}$

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

$x - 5$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{3 x - 1}$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

Substitute $5$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{3 \cdot 5 - 1}$

Simplify the denominator.

Multiply $3$ by $5$ .

$\frac{1}{15 - 1}$

Subtract $1$ from $15$ .

$\frac{1}{14}$

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

$(5, \frac{1}{14})$

Find the Holes in the Graph f(x)=(x-5)/(3x^2-16x+5)

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