Precalculus Examples

$2 - 10 x = (3 x - 7) (x + 1)$

Step 1

Subtract $(3 x - 7) (x + 1)$ from both sides of the equation.

$2 - 10 x - (3 x - 7) (x + 1) = 0$

Step 2

Simplify $2 - 10 x - (3 x - 7) (x + 1)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$2 - 10 x + (- (3 x) - - 7) (x + 1) = 0$

Step 2.1.2

Multiply $3$ by $- 1$ .

$2 - 10 x + (- 3 x - - 7) (x + 1) = 0$

Step 2.1.3

Multiply $- 1$ by $- 7$ .

$2 - 10 x + (- 3 x + 7) (x + 1) = 0$

Step 2.1.4

Expand $(- 3 x + 7) (x + 1)$ using the FOIL Method.

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Step 2.1.4.1

Apply the distributive property.

$2 - 10 x - 3 x (x + 1) + 7 (x + 1) = 0$

Step 2.1.4.2

Apply the distributive property.

$2 - 10 x - 3 x \cdot x - 3 x \cdot 1 + 7 (x + 1) = 0$

Step 2.1.4.3

Apply the distributive property.

$2 - 10 x - 3 x \cdot x - 3 x \cdot 1 + 7 x + 7 \cdot 1 = 0$

Step 2.1.5

Simplify and combine like terms.

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Step 2.1.5.1

Simplify each term.

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Step 2.1.5.1.1

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

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Step 2.1.5.1.1.1

Move $x$ .

$2 - 10 x - 3 (x \cdot x) - 3 x \cdot 1 + 7 x + 7 \cdot 1 = 0$

Step 2.1.5.1.1.2

Multiply $x$ by $x$ .

$2 - 10 x - 3 x^{2} - 3 x \cdot 1 + 7 x + 7 \cdot 1 = 0$

Step 2.1.5.1.2

Multiply $- 3$ by $1$ .

$2 - 10 x - 3 x^{2} - 3 x + 7 x + 7 \cdot 1 = 0$

Step 2.1.5.1.3

Multiply $7$ by $1$ .

$2 - 10 x - 3 x^{2} - 3 x + 7 x + 7 = 0$

Step 2.1.5.2

Add $- 3 x$ and $7 x$ .

$2 - 10 x - 3 x^{2} + 4 x + 7 = 0$

Step 2.2

Add $2$ and $7$ .

$- 10 x - 3 x^{2} + 4 x + 9 = 0$

Step 2.3

Add $- 10 x$ and $4 x$ .

$- 3 x^{2} - 6 x + 9 = 0$

Step 3

Factor $3$ out of $- 3 x^{2} - 6 x + 9$ .

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Step 3.1

Factor $3$ out of $- 3 x^{2}$ .

$3 (- x^{2}) - 6 x + 9 = 0$

Step 3.2

Factor $3$ out of $- 6 x$ .

$3 (- x^{2}) + 3 (- 2 x) + 9 = 0$

Step 3.3

Factor $3$ out of $9$ .

$3 (- x^{2}) + 3 (- 2 x) + 3 (3) = 0$

Step 3.4

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

$3 (- x^{2} - 2 x) + 3 (3) = 0$

Step 3.5

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

$3 (- x^{2} - 2 x + 3) = 0$

Step 4

Factor.

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Step 4.1

Factor by grouping.

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Step 4.1.1

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 $a \cdot c = - 1 \cdot 3 = - 3$ and whose sum is $b = - 2$ .

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Step 4.1.1.1

Factor $- 2$ out of $- 2 x$ .

$3 (- x^{2} - 2 x + 3) = 0$

Step 4.1.1.2

Rewrite $- 2$ as $1$ plus $- 3$

$3 (- x^{2} + (1 - 3) x + 3) = 0$

Step 4.1.1.3

Apply the distributive property.

$3 (- x^{2} + 1 x - 3 x + 3) = 0$

Step 4.1.1.4

Multiply $x$ by $1$ .

$3 (- x^{2} + x - 3 x + 3) = 0$

Step 4.1.2

Factor out the greatest common factor from each group.

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Step 4.1.2.1

Group the first two terms and the last two terms.

$3 ((- x^{2} + x) - 3 x + 3) = 0$

Step 4.1.2.2

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

$3 (x (- x + 1) + 3 (- x + 1)) = 0$

Step 4.1.3

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$3 ((- x + 1) (x + 3)) = 0$

Step 4.2

Remove unnecessary parentheses.

$3 (- x + 1) (x + 3) = 0$

Step 5

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

$- x + 1 = 0$

$x + 3 = 0$

Step 6

Set $- x + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

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

$- x + 1 = 0$

Step 6.2

Solve $- x + 1 = 0$ for $x$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 6.2.2

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

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Step 6.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 7

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

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Step 7.1

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

$x + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8

The final solution is all the values that make $3 (- x + 1) (x + 3) = 0$ true.

$x = 1, - 3$