Algebra Examples

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

Step 1

Factor by grouping.

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Step 1.1

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 $a \cdot c = 3 \cdot - 8 = - 24$ and whose sum is $b = 2$ .

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Step 1.1.1

Factor $2$ out of $2 x$ .

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

Step 1.1.2

Rewrite $2$ as $- 4$ plus $6$

$f (x) = \frac{3 x^{2} + (- 4 + 6) x - 8}{4 - x^{2}}$

Step 1.1.3

Apply the distributive property.

$f (x) = \frac{3 x^{2} - 4 x + 6 x - 8}{4 - x^{2}}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$f (x) = \frac{(3 x^{2} - 4 x) + 6 x - 8}{4 - x^{2}}$

Step 1.2.2

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

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

Step 1.3

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

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

Step 2

Factor $4 - x^{2}$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 3

Cancel the common factor of $x + 2$ and $2 + x$ .

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Step 3.1

Reorder terms.

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

Step 3.2

Cancel the common factor.

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

Step 3.3

Rewrite the expression.

$f (x) = \frac{3 x - 4}{2 - x}$

Step 4

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

$2 + x$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3 x - 4}{2 - x}$ .

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Step 5.1

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

$2 + x = 0$

Step 5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5.3

Substitute $- 2$ for $x$ in $\frac{3 x - 4}{2 - x}$ and simplify.

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Step 5.3.1

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

$\frac{3 \cdot - 2 - 4}{2 - - 2}$

Step 5.3.2

Simplify.

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Step 5.3.2.1

Cancel the common factor of $3 \cdot - 2 - 4$ and $2 - - 2$ .

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Step 5.3.2.1.1

Reorder terms.

$\frac{3 \cdot - 2 - 4}{2 - 2 \cdot - 1}$

Step 5.3.2.1.2

Factor $2$ out of $3 \cdot - 2$ .

$\frac{2 (3 \cdot - 1) - 4}{2 - 2 \cdot - 1}$

Step 5.3.2.1.3

Factor $2$ out of $- 4$ .

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

Step 5.3.2.1.4

Factor $2$ out of $2 (3 \cdot - 1) + 2 (- 2)$ .

$\frac{2 (3 \cdot - 1 - 2)}{2 - 2 \cdot - 1}$

Step 5.3.2.1.5

Cancel the common factors.

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Step 5.3.2.1.5.1

Factor $2$ out of $2$ .

$\frac{2 (3 \cdot - 1 - 2)}{2 (1) - 2 \cdot - 1}$

Step 5.3.2.1.5.2

Factor $2$ out of $- 2 \cdot - 1$ .

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

Step 5.3.2.1.5.3

Factor $2$ out of $2 (1) + 2 (- - 1)$ .

$\frac{2 (3 \cdot - 1 - 2)}{2 (1 - - 1)}$

Step 5.3.2.1.5.4

Cancel the common factor.

$\frac{2 (3 \cdot - 1 - 2)}{2 (1 - - 1)}$

Step 5.3.2.1.5.5

Rewrite the expression.

$\frac{3 \cdot - 1 - 2}{1 - - 1}$

Step 5.3.2.2

Simplify the numerator.

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Step 5.3.2.2.1

Multiply $3$ by $- 1$ .

$\frac{- 3 - 2}{1 - - 1}$

Step 5.3.2.2.2

Subtract $2$ from $- 3$ .

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

Step 5.3.2.3

Simplify the denominator.

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Step 5.3.2.3.1

Multiply $- 1$ by $- 1$ .

$\frac{- 5}{1 + 1}$

Step 5.3.2.3.2

Add $1$ and $1$ .

$\frac{- 5}{2}$

Step 5.3.2.4

Move the negative in front of the fraction.

$- \frac{5}{2}$

Step 5.4

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

$(- 2, - \frac{5}{2})$

Step 6