Find the Holes in the Graph (3x^2-5x-12)/(x^2-4x+3)

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

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = - 5$ .

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

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

Rewrite $- 5$ as $4$ plus $- 9$

$\frac{3 x^{2} + (4 - 9) x - 12}{x^{2} - 4 x + 3}$

Apply the distributive property.

$\frac{3 x^{2} + 4 x - 9 x - 12}{x^{2} - 4 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(3 x^{2} + 4 x) - 9 x - 12}{x^{2} - 4 x + 3}$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

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

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.

$\frac{(3 x + 4) (x - 3)}{(x - 3) (x - 1)}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(3 x + 4) (x - 3)}{(x - 3) (x - 1)}$

Rewrite the expression.

$\frac{3 x + 4}{x - 1}$

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 $\frac{3 x + 4}{x - 1}$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{3 x + 4}{x - 1}$ and simplify.

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

$\frac{3 \cdot 3 + 4}{3 - 1}$

Simplify.

Simplify the numerator.

Multiply $3$ by $3$ .

$\frac{9 + 4}{3 - 1}$

Add $9$ and $4$ .

$\frac{13}{3 - 1}$

Subtract $1$ from $3$ .

$\frac{13}{2}$

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

$(3, \frac{13}{2})$

Find the Holes in the Graph (3x^2-5x-12)/(x^2-4x+3)

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