Algebra Examples

$f (x) = \frac{(x^{2} + x) (x^{2} - 8 x + 16)}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1

Factor $(x^{2} + x) (x^{2} - 8 x + 16)$ .

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Step 1.1

Factor $x$ out of $x^{2} + x$ .

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Step 1.1.1

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

$f (x) = \frac{(x \cdot x + x) (x^{2} - 8 x + 16)}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.1.2

Raise $x$ to the power of $1$ .

$f (x) = \frac{(x \cdot x + x) (x^{2} - 8 x + 16)}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.1.3

Factor $x$ out of $x^{1}$ .

$f (x) = \frac{(x \cdot x + x \cdot 1) (x^{2} - 8 x + 16)}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.1.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$f (x) = \frac{x (x + 1) (x^{2} - 8 x + 16)}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.2

Factor using the perfect square rule.

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Step 1.2.1

Rewrite $16$ as $4^{2}$ .

$f (x) = \frac{x (x + 1) (x^{2} - 8 x + 4^{2})}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Step 1.2.3

Rewrite the polynomial.

$f (x) = \frac{x (x + 1) (x^{2} - 2 \cdot x \cdot 4 + 4^{2})}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x^{2} - 1) (10 x^{3} - 15 x)}$

Step 2

Factor $(x^{2} - 1) (10 x^{3} - 15 x)$ .

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Step 2.1

Rewrite $1$ as $1^{2}$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x^{2} - 1^{2}) (10 x^{3} - 15 x)}$

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 $a = x$ and $b = 1$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) (10 x^{3} - 15 x)}$

Step 2.3

Factor.

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Step 2.3.1

Factor $5 x$ out of $10 x^{3} - 15 x$ .

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Step 2.3.1.1

Factor $5 x$ out of $10 x^{3}$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) (5 x (2 x^{2}) - 15 x)}$

Step 2.3.1.2

Factor $5 x$ out of $- 15 x$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) (5 x (2 x^{2}) + 5 x (- 3))}$

Step 2.3.1.3

Factor $5 x$ out of $5 x (2 x^{2}) + 5 x (- 3)$ .

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) (5 x (2 x^{2} - 3))}$

Step 2.3.2

Remove unnecessary parentheses.

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) \cdot (5 x (2 x^{2} - 3))}$

Step 3

Cancel the common factor of $x$ .

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Step 3.1

Cancel the common factor.

$f (x) = \frac{x (x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) \cdot (5 x (2 x^{2} - 3))}$

Step 3.2

Rewrite the expression.

$f (x) = \frac{(x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) \cdot (5 (2 x^{2} - 3))}$

Step 4

Cancel the common factor of $x + 1$ .

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Step 4.1

Cancel the common factor.

$f (x) = \frac{(x + 1) {(x - 4)}^{2}}{(x + 1) (x - 1) \cdot (5 (2 x^{2} - 3))}$

Step 4.2

Rewrite the expression.

$f (x) = \frac{{(x - 4)}^{2}}{(x - 1) \cdot (5 (2 x^{2} - 3))}$

Step 5

Move $5$ to the left of $x - 1$ .

$f (x) = \frac{{(x - 4)}^{2}}{5 \cdot ((x - 1) (2 x^{2} - 3))}$

Step 6

Multiply $5$ by $x - 1$ .

$f (x) = \frac{{(x - 4)}^{2}}{5 (x - 1) (2 x^{2} - 3)}$

Step 7

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

$x + 1, x$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{{(x - 4)}^{2}}{5 (x - 1) (2 x^{2} - 3)}$ .

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Step 8.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8.3

Substitute $- 1$ for $x$ in $\frac{{(x - 4)}^{2}}{5 (x - 1) (2 x^{2} - 3)}$ and simplify.

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Step 8.3.1

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

$\frac{{(- 1 - 4)}^{2}}{5 (- 1 - 1) (2 {(- 1)}^{2} - 3)}$

Step 8.3.2

Simplify.

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Step 8.3.2.1

Simplify the numerator.

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Step 8.3.2.1.1

Subtract $4$ from $- 1$ .

$\frac{{(- 5)}^{2}}{5 (- 1 - 1) (2 {(- 1)}^{2} - 3)}$

Step 8.3.2.1.2

Raise $- 5$ to the power of $2$ .

$\frac{25}{5 (- 1 - 1) (2 {(- 1)}^{2} - 3)}$

Step 8.3.2.2

Simplify the denominator.

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Step 8.3.2.2.1

Subtract $1$ from $- 1$ .

$\frac{25}{5 \cdot - 2 (2 {(- 1)}^{2} - 3)}$

Step 8.3.2.2.2

Multiply $5$ by $- 2$ .

$\frac{25}{- 10 (2 {(- 1)}^{2} - 3)}$

Step 8.3.2.2.3

Raise $- 1$ to the power of $2$ .

$\frac{25}{- 10 (2 \cdot 1 - 3)}$

Step 8.3.2.2.4

Multiply $2$ by $1$ .

$\frac{25}{- 10 (2 - 3)}$

Step 8.3.2.2.5

Subtract $3$ from $2$ .

$\frac{25}{- 10 \cdot - 1}$

Step 8.3.2.3

Multiply $- 10$ by $- 1$ .

$\frac{25}{10}$

Step 8.3.2.4

Cancel the common factor of $25$ and $10$ .

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Step 8.3.2.4.1

Factor $5$ out of $25$ .

$\frac{5 (5)}{10}$

Step 8.3.2.4.2

Cancel the common factors.

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Step 8.3.2.4.2.1

Factor $5$ out of $10$ .

$\frac{5 \cdot 5}{5 \cdot 2}$

Step 8.3.2.4.2.2

Cancel the common factor.

$\frac{5 \cdot 5}{5 \cdot 2}$

Step 8.3.2.4.2.3

Rewrite the expression.

$\frac{5}{2}$

Step 8.4

Set $x$ equal to $0$ .

$x = 0$

Step 8.5

Substitute $0$ for $x$ in $\frac{{(x - 4)}^{2}}{5 (x - 1) (2 x^{2} - 3)}$ and simplify.

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Step 8.5.1

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{{(0 - 4)}^{2}}{5 (0 - 1) (2 \cdot 0^{2} - 3)}$

Step 8.5.2

Simplify.

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Step 8.5.2.1

Simplify the numerator.

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Step 8.5.2.1.1

Subtract $4$ from $0$ .

$\frac{{(- 4)}^{2}}{5 (0 - 1) (2 \cdot 0^{2} - 3)}$

Step 8.5.2.1.2

Raise $- 4$ to the power of $2$ .

$\frac{16}{5 (0 - 1) (2 \cdot 0^{2} - 3)}$

Step 8.5.2.2

Simplify the denominator.

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Step 8.5.2.2.1

Subtract $1$ from $0$ .

$\frac{16}{5 \cdot - 1 (2 \cdot 0^{2} - 3)}$

Step 8.5.2.2.2

Combine exponents.

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Step 8.5.2.2.2.1

Factor out negative.

$\frac{16}{- (5 (2 \cdot 0^{2} - 3))}$

Step 8.5.2.2.2.2

Multiply $5$ by $- 1$ .

$\frac{16}{- 5 (2 \cdot 0^{2} - 3)}$

Step 8.5.2.2.3

Simplify each term.

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Step 8.5.2.2.3.1

Raising $0$ to any positive power yields $0$ .

$\frac{16}{- 5 (2 \cdot 0 - 3)}$

Step 8.5.2.2.3.2

Multiply $2$ by $0$ .

$\frac{16}{- 5 (0 - 3)}$

Step 8.5.2.2.4

Subtract $3$ from $0$ .

$\frac{16}{- 5 \cdot - 3}$

Step 8.5.2.3

Multiply $- 5$ by $- 3$ .

$\frac{16}{15}$

Step 8.6

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

$(- 1, \frac{5}{2}), (0, \frac{16}{15})$

Step 9