Find the Holes in the Graph (-3x^2+21x-36)/(5x-20)

$\frac{- 3 x^{2} + 21 x - 36}{5 x - 20}$

Step 1

Factor $- 3 x^{2} + 21 x - 36$ .

Factor $3$ out of $- 3 x^{2} + 21 x - 36$ .

Factor $3$ out of $- 3 x^{2}$ .

$\frac{3 (- x^{2}) + 21 x - 36}{5 x - 20}$

Factor $3$ out of $21 x$ .

$\frac{3 (- x^{2}) + 3 (7 x) - 36}{5 x - 20}$

Factor $3$ out of $- 36$ .

$\frac{3 (- x^{2}) + 3 (7 x) + 3 (- 12)}{5 x - 20}$

Factor $3$ out of $3 (- x^{2}) + 3 (7 x)$ .

$\frac{3 (- x^{2} + 7 x) + 3 (- 12)}{5 x - 20}$

Factor $3$ out of $3 (- x^{2} + 7 x) + 3 (- 12)$ .

$\frac{3 (- x^{2} + 7 x - 12)}{5 x - 20}$

Factor.

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 $a \cdot c = - 1 \cdot - 12 = 12$ and whose sum is $b = 7$ .

Factor $7$ out of $7 x$ .

$\frac{3 (- x^{2} + 7 (x) - 12)}{5 x - 20}$

Rewrite $7$ as $3$ plus $4$

$\frac{3 (- x^{2} + (3 + 4) x - 12)}{5 x - 20}$

Apply the distributive property.

$\frac{3 (- x^{2} + 3 x + 4 x - 12)}{5 x - 20}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{3 ((- x^{2} + 3 x) + 4 x - 12)}{5 x - 20}$

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

$\frac{3 (x (- x + 3) - 4 (- x + 3))}{5 x - 20}$

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

$\frac{3 ((- x + 3) (x - 4))}{5 x - 20}$

Remove unnecessary parentheses.

$\frac{3 (- x + 3) (x - 4)}{5 x - 20}$

Step 2

Factor $5$ out of $5 x - 20$ .

Factor $5$ out of $5 x$ .

$\frac{3 (- x + 3) (x - 4)}{5 (x) - 20}$

Factor $5$ out of $- 20$ .

$\frac{3 (- x + 3) (x - 4)}{5 x + 5 \cdot - 4}$

Factor $5$ out of $5 x + 5 \cdot - 4$ .

$\frac{3 (- x + 3) (x - 4)}{5 (x - 4)}$

Step 3

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{3 (- x + 3) (x - 4)}{5 (x - 4)}$

Rewrite the expression.

$\frac{3 (- x + 3)}{5}$

Step 4

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

$x - 4$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 (- x + 3)}{5}$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{3 (- x + 3)}{5}$ and simplify.

Substitute $4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 (- 4 + 3)}{5}$

Simplify.

Add $- 4$ and $3$ .

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

Multiply $3$ by $- 1$ .

$\frac{- 3}{5}$

Move the negative in front of the fraction.

$- \frac{3}{5}$

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

$(4, - \frac{3}{5})$

Step 6

Find the Holes in the Graph (-3x^2+21x-36)/(5x-20)

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