Algebra Examples

${(2 x + 1)}^{2} = 8 x + 25$

Step 1

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

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Step 1.1

Subtract $8 x$ from both sides of the equation.

${(2 x + 1)}^{2} - 8 x = 25$

Step 1.2

Simplify each term.

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Step 1.2.1

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

$(2 x + 1) (2 x + 1) - 8 x = 25$

Step 1.2.2

Expand $(2 x + 1) (2 x + 1)$ using the FOIL Method.

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Step 1.2.2.1

Apply the distributive property.

$2 x (2 x + 1) + 1 (2 x + 1) - 8 x = 25$

Step 1.2.2.2

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 1 + 1 (2 x + 1) - 8 x = 25$

Step 1.2.2.3

Apply the distributive property.

$2 x (2 x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3

Simplify and combine like terms.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.3.1.2.1

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.2.2

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.3

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.4

Multiply $2$ by $1$ .

$4 x^{2} + 2 x + 1 (2 x) + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.5

Multiply $2 x$ by $1$ .

$4 x^{2} + 2 x + 2 x + 1 \cdot 1 - 8 x = 25$

Step 1.2.3.1.6

Multiply $1$ by $1$ .

$4 x^{2} + 2 x + 2 x + 1 - 8 x = 25$

Step 1.2.3.2

Add $2 x$ and $2 x$ .

$4 x^{2} + 4 x + 1 - 8 x = 25$

Step 1.3

Subtract $8 x$ from $4 x$ .

$4 x^{2} - 4 x + 1 = 25$

Step 2

Subtract $25$ from both sides of the equation.

$4 x^{2} - 4 x + 1 - 25 = 0$

Step 3

Subtract $25$ from $1$ .

$4 x^{2} - 4 x - 24 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $4$ out of $4 x^{2} - 4 x - 24$ .

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Step 4.1.1

Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) - 4 x - 24 = 0$

Step 4.1.2

Factor $4$ out of $- 4 x$ .

$4 (x^{2}) + 4 (- x) - 24 = 0$

Step 4.1.3

Factor $4$ out of $- 24$ .

$4 (x^{2}) + 4 (- x) + 4 (- 6) = 0$

Step 4.1.4

Factor $4$ out of $4 (x^{2}) + 4 (- x)$ .

$4 (x^{2} - x) + 4 (- 6) = 0$

Step 4.1.5

Factor $4$ out of $4 (x^{2} - x) + 4 (- 6)$ .

$4 (x^{2} - x - 6) = 0$

Step 4.2

Factor.

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Step 4.2.1

Factor $x^{2} - x - 6$ using the AC method.

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Step 4.2.1.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 4.2.1.2

Write the factored form using these integers.

$4 ((x - 3) (x + 2)) = 0$

Step 4.2.2

Remove unnecessary parentheses.

$4 (x - 3) (x + 2) = 0$

Step 5

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

$x - 3 = 0$

$x + 2 = 0$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

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

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

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

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Step 7.1

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

$x + 2 = 0$

Step 7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8

The final solution is all the values that make $4 (x - 3) (x + 2) = 0$ true.

$x = 3, - 2$