Precalculus Examples

${(1 + y)}^{2} - (1 + y) y - y^{2} + 1 = 0$

Step 1

Simplify ${(1 + y)}^{2} - (1 + y) y - y^{2} + 1$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

${(1 + y)}^{2} + (- 1 \cdot 1 - y) y - y^{2} + 1 = 0$

Step 1.1.2

Multiply $- 1$ by $1$ .

${(1 + y)}^{2} + (- 1 - y) y - y^{2} + 1 = 0$

Step 1.1.3

Apply the distributive property.

${(1 + y)}^{2} - 1 y - y \cdot y - y^{2} + 1 = 0$

Step 1.1.4

Rewrite $- 1 y$ as $- y$ .

${(1 + y)}^{2} - y - y \cdot y - y^{2} + 1 = 0$

Step 1.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 1.1.5.1

Move $y$ .

${(1 + y)}^{2} - y - (y \cdot y) - y^{2} + 1 = 0$

Step 1.1.5.2

Multiply $y$ by $y$ .

${(1 + y)}^{2} - y - y^{2} - y^{2} + 1 = 0$

Step 1.2

Subtract $y^{2}$ from $- y^{2}$ .

${(1 + y)}^{2} - y - 2 y^{2} + 1 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Regroup terms.

${(1 + y)}^{2} - 2 y^{2} - y + 1 = 0$

Step 2.2

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

$(1 + y) (1 + y) - 2 y^{2} - y + 1 = 0$

Step 2.3

Expand $(1 + y) (1 + y)$ using the FOIL Method.

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Step 2.3.1

Apply the distributive property.

$1 (1 + y) + y (1 + y) - 2 y^{2} - y + 1 = 0$

Step 2.3.2

Apply the distributive property.

$1 \cdot 1 + 1 y + y (1 + y) - 2 y^{2} - y + 1 = 0$

Step 2.3.3

Apply the distributive property.

$1 \cdot 1 + 1 y + y \cdot 1 + y \cdot y - 2 y^{2} - y + 1 = 0$

Step 2.4

Simplify and combine like terms.

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Step 2.4.1

Simplify each term.

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Step 2.4.1.1

Multiply $1$ by $1$ .

$1 + 1 y + y \cdot 1 + y \cdot y - 2 y^{2} - y + 1 = 0$

Step 2.4.1.2

Multiply $y$ by $1$ .

$1 + y + y \cdot 1 + y \cdot y - 2 y^{2} - y + 1 = 0$

Step 2.4.1.3

Multiply $y$ by $1$ .

$1 + y + y + y \cdot y - 2 y^{2} - y + 1 = 0$

Step 2.4.1.4

Multiply $y$ by $y$ .

$1 + y + y + y^{2} - 2 y^{2} - y + 1 = 0$

Step 2.4.2

Add $y$ and $y$ .

$1 + 2 y + y^{2} - 2 y^{2} - y + 1 = 0$

Step 2.5

Subtract $2 y^{2}$ from $y^{2}$ .

$1 + 2 y - y^{2} - y + 1 = 0$

Step 2.6

Factor $- 1$ out of $1 + 2 y - y^{2}$ .

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Step 2.6.1

Reorder the expression.

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Step 2.6.1.1

Move $1$ .

$2 y - y^{2} + 1 - y + 1 = 0$

Step 2.6.1.2

Reorder $2 y$ and $- y^{2}$ .

$- y^{2} + 2 y + 1 - y + 1 = 0$

Step 2.6.2

Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) + 2 y + 1 - y + 1 = 0$

Step 2.6.3

Factor $- 1$ out of $2 y$ .

$- (y^{2}) - (- 2 y) + 1 - y + 1 = 0$

Step 2.6.4

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

$- (y^{2}) - (- 2 y) - 1 \cdot - 1 - y + 1 = 0$

Step 2.6.5

Factor $- 1$ out of $- (y^{2}) - (- 2 y)$ .

$- (y^{2} - 2 y) - 1 \cdot - 1 - y + 1 = 0$

Step 2.6.6

Factor $- 1$ out of $- (y^{2} - 2 y) - 1 (- 1)$ .

$- (y^{2} - 2 y - 1) - y + 1 = 0$

Step 2.7

Factor $- 1$ out of $- y + 1$ .

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Step 2.7.1

Factor $- 1$ out of $- y$ .

$- (y^{2} - 2 y - 1) - (y) + 1 = 0$

Step 2.7.2

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

$- (y^{2} - 2 y - 1) - (y) - 1 \cdot - 1 = 0$

Step 2.7.3

Factor $- 1$ out of $- (y) - 1 (- 1)$ .

$- (y^{2} - 2 y - 1) - (y - 1) = 0$

Step 2.8

Factor $- 1$ out of $- (y^{2} - 2 y - 1) - (y - 1)$ .

$- (y^{2} - 2 y - 1 + y - 1) = 0$

Step 2.9

Add $- 2 y$ and $y$ .

$- (y^{2} - y - 1 - 1) = 0$

Step 2.10

Subtract $1$ from $- 1$ .

$- (y^{2} - y - 2) = 0$

Step 2.11

Factor.

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Step 2.11.1

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

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Step 2.11.1.1

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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 2.11.1.2

Write the factored form using these integers.

$- ((y - 2) (y + 1)) = 0$

Step 2.11.2

Remove unnecessary parentheses.

$- (y - 2) (y + 1) = 0$

Step 3

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

$y - 2 = 0$

$y + 1 = 0$

Step 4

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

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Step 4.1

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

$y - 2 = 0$

Step 4.2

Add $2$ to both sides of the equation.

$y = 2$

Step 5

Set $y + 1$ equal to $0$ and solve for $y$ .

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Step 5.1

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

$y + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 6

The final solution is all the values that make $- (y - 2) (y + 1) = 0$ true.

$y = 2, - 1$