Basic Math Examples

$\frac{1}{k + 2} - \frac{4}{k - 2} = \frac{k^{2}}{4 - k^{2}}$

Step 1

Factor each term.

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Step 1.1

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

$\frac{1}{k + 2} - \frac{4}{k - 2} = \frac{k^{2}}{2^{2} - k^{2}}$

Step 1.2

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 $b = k$ .

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

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$k + 2, k - 2, (2 + k) (2 - k)$

Step 2.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.5

The factor for $k + 2$ is $k + 2$ itself.

$(k + 2) = k + 2$

$(k + 2)$ occurs $1$ time.

Step 2.6

The factor for $k - 2$ is $k - 2$ itself.

$(k - 2) = k - 2$

$(k - 2)$ occurs $1$ time.

Step 2.7

The factor for $2 + k$ is $2 + k$ itself.

$(2 + k) = 2 + k$

$(2 + k)$ occurs $1$ time.

Step 2.8

The factor for $2 - k$ is $2 - k$ itself.

$(2 - k) = 2 - k$

$(2 - k)$ occurs $1$ time.

Step 2.9

The LCM of $k + 2, k - 2, 2 + k, 2 - k$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(k + 2) (k - 2) (2 - k)$

Step 3

Multiply each term in $\frac{1}{k + 2} - \frac{4}{k - 2} = \frac{k^{2}}{(2 + k) (2 - k)}$ by $(k + 2) (k - 2) (2 - k)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{1}{k + 2} - \frac{4}{k - 2} = \frac{k^{2}}{(2 + k) (2 - k)}$ by $(k + 2) (k - 2) (2 - k)$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $k + 2$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor.

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

Step 3.2.1.1.3

Rewrite the expression.

$(k - 2) (2 - k) - \frac{4}{k - 2} ((k + 2) (k - 2) (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.2

Factor $- 1$ out of $k$ .

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

Step 3.2.1.3

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 3.2.1.4

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

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

Step 3.2.1.5

Reorder terms.

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

Step 3.2.1.6

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

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

Step 3.2.1.7

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

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

Step 3.2.1.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.2.1.9

Add $1$ and $1$ .

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

Step 3.2.1.10

Rewrite $- 1 {(- k + 2)}^{2}$ as $- {(- k + 2)}^{2}$ .

$- {(- k + 2)}^{2} - \frac{4}{k - 2} ((k + 2) (k - 2) (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.11

Cancel the common factor of $k - 2$ .

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Step 3.2.1.11.1

Move the leading negative in $- \frac{4}{k - 2}$ into the numerator.

$- {(- k + 2)}^{2} + \frac{- 4}{k - 2} ((k + 2) (k - 2) (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.11.2

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

$- {(- k + 2)}^{2} + \frac{- 4}{k - 2} ((k - 2) ((k + 2) (2 - k))) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.11.3

Cancel the common factor.

$- {(- k + 2)}^{2} + \frac{- 4}{k - 2} ((k - 2) ((k + 2) (2 - k))) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.11.4

Rewrite the expression.

$- {(- k + 2)}^{2} - 4 ((k + 2) (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.12

Expand $(k + 2) (2 - k)$ using the FOIL Method.

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Step 3.2.1.12.1

Apply the distributive property.

$- {(- k + 2)}^{2} - 4 (k (2 - k) + 2 (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.12.2

Apply the distributive property.

$- {(- k + 2)}^{2} - 4 (k \cdot 2 + k (- k) + 2 (2 - k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.12.3

Apply the distributive property.

$- {(- k + 2)}^{2} - 4 (k \cdot 2 + k (- k) + 2 \cdot 2 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13

Simplify and combine like terms.

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Step 3.2.1.13.1

Simplify each term.

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Step 3.2.1.13.1.1

Move $2$ to the left of $k$ .

$- {(- k + 2)}^{2} - 4 (2 \cdot k + k (- k) + 2 \cdot 2 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.1.2

Rewrite using the commutative property of multiplication.

$- {(- k + 2)}^{2} - 4 (2 k - k \cdot k + 2 \cdot 2 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.1.3

Multiply $k$ by $k$ by adding the exponents.

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Step 3.2.1.13.1.3.1

Move $k$ .

$- {(- k + 2)}^{2} - 4 (2 k - (k \cdot k) + 2 \cdot 2 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.1.3.2

Multiply $k$ by $k$ .

$- {(- k + 2)}^{2} - 4 (2 k - k^{2} + 2 \cdot 2 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.1.4

Multiply $2$ by $2$ .

$- {(- k + 2)}^{2} - 4 (2 k - k^{2} + 4 + 2 (- k)) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.1.5

Multiply $- 1$ by $2$ .

$- {(- k + 2)}^{2} - 4 (2 k - k^{2} + 4 - 2 k) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.2

Subtract $2 k$ from $2 k$ .

$- {(- k + 2)}^{2} - 4 (- k^{2} + 4 + 0) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.13.3

Add $- k^{2}$ and $0$ .

$- {(- k + 2)}^{2} - 4 (- k^{2} + 4) = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.14

Apply the distributive property.

$- {(- k + 2)}^{2} - 4 (- k^{2}) - 4 \cdot 4 = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.15

Multiply $- 1$ by $- 4$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 4 \cdot 4 = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.2.1.16

Multiply $- 4$ by $4$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{(2 + k) (2 - k)} ((k + 2) (k - 2) (2 - k))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $2 - k$ .

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Step 3.3.1.1

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

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{(2 - k) (2 + k)} ((k + 2) (k - 2) (2 - k))$

Step 3.3.1.2

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

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{(2 - k) (2 + k)} ((2 - k) ((k + 2) (k - 2)))$

Step 3.3.1.3

Cancel the common factor.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{(2 - k) (2 + k)} ((2 - k) ((k + 2) (k - 2)))$

Step 3.3.1.4

Rewrite the expression.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} ((k + 2) (k - 2))$

Step 3.3.2

Expand $(k + 2) (k - 2)$ using the FOIL Method.

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Step 3.3.2.1

Apply the distributive property.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k (k - 2) + 2 (k - 2))$

Step 3.3.2.2

Apply the distributive property.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k \cdot k + k \cdot - 2 + 2 (k - 2))$

Step 3.3.2.3

Apply the distributive property.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k \cdot k + k \cdot - 2 + 2 k + 2 \cdot - 2)$

Step 3.3.3

Simplify terms.

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Step 3.3.3.1

Combine the opposite terms in $k \cdot k + k \cdot - 2 + 2 k + 2 \cdot - 2$ .

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Step 3.3.3.1.1

Reorder the factors in the terms $k \cdot - 2$ and $2 k$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k \cdot k - 2 k + 2 k + 2 \cdot - 2)$

Step 3.3.3.1.2

Add $- 2 k$ and $2 k$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k \cdot k + 0 + 2 \cdot - 2)$

Step 3.3.3.1.3

Add $k \cdot k$ and $0$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k \cdot k + 2 \cdot - 2)$

Step 3.3.3.2

Simplify each term.

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Step 3.3.3.2.1

Multiply $k$ by $k$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k^{2} + 2 \cdot - 2)$

Step 3.3.3.2.2

Multiply $2$ by $- 2$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2}}{2 + k} (k^{2} - 4)$

Step 3.3.3.3

Multiply $\frac{k^{2}}{2 + k}$ by $k^{2} - 4$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2} (k^{2} - 4)}{2 + k}$

Step 3.3.4

Simplify the numerator.

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Step 3.3.4.1

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

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2} (k^{2} - 2^{2})}{2 + k}$

Step 3.3.4.2

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

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2} (k + 2) (k - 2)}{2 + k}$

Step 3.3.5

Simplify terms.

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Step 3.3.5.1

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

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Step 3.3.5.1.1

Reorder terms.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2} (k + 2) (k - 2)}{k + 2}$

Step 3.3.5.1.2

Cancel the common factor.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = \frac{k^{2} (k + 2) (k - 2)}{k + 2}$

Step 3.3.5.1.3

Divide $k^{2} (k - 2)$ by $1$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{2} (k - 2)$

Step 3.3.5.2

Apply the distributive property.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{2} k + k^{2} \cdot - 2$

Step 3.3.6

Multiply $k^{2}$ by $k$ by adding the exponents.

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Step 3.3.6.1

Multiply $k^{2}$ by $k$ .

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Step 3.3.6.1.1

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

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{2} k^{1} + k^{2} \cdot - 2$

Step 3.3.6.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{2 + 1} + k^{2} \cdot - 2$

Step 3.3.6.2

Add $2$ and $1$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{3} + k^{2} \cdot - 2$

Step 3.3.7

Move $- 2$ to the left of $k^{2}$ .

$- {(- k + 2)}^{2} + 4 k^{2} - 16 = k^{3} - 2 k^{2}$

Step 4

Solve the equation.

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Step 4.1

Move all terms containing $k$ to the left side of the equation.

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Step 4.1.1

Subtract $k^{3}$ from both sides of the equation.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 - k^{3} = - 2 k^{2}$

Step 4.1.2

Add $2 k^{2}$ to both sides of the equation.

$- {(- k + 2)}^{2} + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3

Simplify each term.

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Step 4.1.3.1

Rewrite ${(- k + 2)}^{2}$ as $(- k + 2) (- k + 2)$ .

$- ((- k + 2) (- k + 2)) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.2

Expand $(- k + 2) (- k + 2)$ using the FOIL Method.

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Step 4.1.3.2.1

Apply the distributive property.

$- (- k (- k + 2) + 2 (- k + 2)) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.2.2

Apply the distributive property.

$- (- k (- k) - k \cdot 2 + 2 (- k + 2)) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.2.3

Apply the distributive property.

$- (- k (- k) - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3

Simplify and combine like terms.

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Step 4.1.3.3.1

Simplify each term.

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Step 4.1.3.3.1.1

Rewrite using the commutative property of multiplication.

$- (- 1 \cdot - 1 k \cdot k - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.2

Multiply $k$ by $k$ by adding the exponents.

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Step 4.1.3.3.1.2.1

Move $k$ .

$- (- 1 \cdot - 1 (k \cdot k) - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.2.2

Multiply $k$ by $k$ .

$- (- 1 \cdot - 1 k^{2} - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.3

Multiply $- 1$ by $- 1$ .

$- (1 k^{2} - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.4

Multiply $k^{2}$ by $1$ .

$- (k^{2} - k \cdot 2 + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.5

Multiply $2$ by $- 1$ .

$- (k^{2} - 2 k + 2 (- k) + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.6

Multiply $- 1$ by $2$ .

$- (k^{2} - 2 k - 2 k + 2 \cdot 2) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.1.7

Multiply $2$ by $2$ .

$- (k^{2} - 2 k - 2 k + 4) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.3.2

Subtract $2 k$ from $- 2 k$ .

$- (k^{2} - 4 k + 4) + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.4

Apply the distributive property.

$- k^{2} - (- 4 k) - 1 \cdot 4 + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.5

Simplify.

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Step 4.1.3.5.1

Multiply $- 4$ by $- 1$ .

$- k^{2} + 4 k - 1 \cdot 4 + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.3.5.2

Multiply $- 1$ by $4$ .

$- k^{2} + 4 k - 4 + 4 k^{2} - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.4

Add $- k^{2}$ and $4 k^{2}$ .

$3 k^{2} + 4 k - 4 - 16 - k^{3} + 2 k^{2} = 0$

Step 4.1.5

Add $3 k^{2}$ and $2 k^{2}$ .

$5 k^{2} + 4 k - 4 - 16 - k^{3} = 0$

Step 4.1.6

Subtract $16$ from $- 4$ .

$5 k^{2} + 4 k - 20 - k^{3} = 0$

Step 4.2

Factor the left side of the equation.

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Step 4.2.1

Reorder terms.

$- k^{3} + 5 k^{2} + 4 k - 20 = 0$

Step 4.2.2

Factor out the greatest common factor from each group.

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Step 4.2.2.1

Group the first two terms and the last two terms.

$(- k^{3} + 5 k^{2}) + 4 k - 20 = 0$

Step 4.2.2.2

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

$k^{2} (- k + 5) - 4 (- k + 5) = 0$

Step 4.2.3

Factor the polynomial by factoring out the greatest common factor, $- k + 5$ .

$(- k + 5) (k^{2} - 4) = 0$

Step 4.2.4

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

$(- k + 5) (k^{2} - 2^{2}) = 0$

Step 4.2.5

Factor.

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Step 4.2.5.1

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

$(- k + 5) ((k + 2) (k - 2)) = 0$

Step 4.2.5.2

Remove unnecessary parentheses.

$(- k + 5) (k + 2) (k - 2) = 0$

Step 4.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$- k + 5 = 0$

$k + 2 = 0$

$k - 2 = 0$

Step 4.4

Set $- k + 5$ equal to $0$ and solve for $k$ .

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Step 4.4.1

Set $- k + 5$ equal to $0$ .

$- k + 5 = 0$

Step 4.4.2

Solve $- k + 5 = 0$ for $k$ .

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Step 4.4.2.1

Subtract $5$ from both sides of the equation.

$- k = - 5$

Step 4.4.2.2

Divide each term in $- k = - 5$ by $- 1$ and simplify.

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Step 4.4.2.2.1

Divide each term in $- k = - 5$ by $- 1$ .

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

Step 4.4.2.2.2

Simplify the left side.

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Step 4.4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.4.2.2.2.2

Divide $k$ by $1$ .

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

Step 4.4.2.2.3

Simplify the right side.

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Step 4.4.2.2.3.1

Divide $- 5$ by $- 1$ .

$k = 5$

Step 4.5

Set $k + 2$ equal to $0$ and solve for $k$ .

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Step 4.5.1

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

$k + 2 = 0$

Step 4.5.2

Subtract $2$ from both sides of the equation.

$k = - 2$

Step 4.6

Set $k - 2$ equal to $0$ and solve for $k$ .

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Step 4.6.1

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

$k - 2 = 0$

Step 4.6.2

Add $2$ to both sides of the equation.

$k = 2$

Step 4.7

The final solution is all the values that make $(- k + 5) (k + 2) (k - 2) = 0$ true.

$k = 5, - 2, 2$

Step 5

Exclude the solutions that do not make $\frac{1}{k + 2} - \frac{4}{k - 2} = \frac{k^{2}}{4 - k^{2}}$ true.

$k = 5$