Find the Holes in the Graph (2x^3+x^2-8x-4)/(x^2-3x+2)

$\frac{2 x^{3} + x^{2} - 8 x - 4}{x^{2} - 3 x + 2}$

Step 1

Factor $2 x^{3} + x^{2} - 8 x - 4$ .

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(2 x^{3} + x^{2}) - 8 x - 4}{x^{2} - 3 x + 2}$

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

$\frac{x^{2} (2 x + 1) - 4 (2 x + 1)}{x^{2} - 3 x + 2}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$\frac{(2 x + 1) (x^{2} - 4)}{x^{2} - 3 x + 2}$

Rewrite $4$ as $2^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$- 2, - 1$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 2$

Step 5

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

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $\frac{(2 x + 1) (x + 2)}{x - 1}$ and simplify.

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{(2 \cdot 2 + 1) (2 + 2)}{2 - 1}$

Simplify.

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{(4 + 1) (2 + 2)}{2 - 1}$

Add $4$ and $1$ .

$\frac{5 (2 + 2)}{2 - 1}$

Add $2$ and $2$ .

$\frac{5 \cdot 4}{2 - 1}$

Subtract $1$ from $2$ .

$\frac{5 \cdot 4}{1}$

Multiply $5$ by $4$ .

$\frac{20}{1}$

Divide $20$ by $1$ .

$20$

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

$(2, 20)$

Step 6

Find the Holes in the Graph (2x^3+x^2-8x-4)/(x^2-3x+2)

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