Solve by Factoring 7-7x=(3x+2)(x-1)

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

Step 1

Subtract $(3 x + 2) (x - 1)$ from both sides of the equation.

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

Step 2

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

Simplify each term.

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Expand $(- 3 x - 2) (x - 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 3$ .

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

Multiply $- 2$ by $- 1$ .

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

Subtract $2 x$ from $3 x$ .

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

Add $7$ and $2$ .

$- 7 x - 3 x^{2} + x + 9 = 0$

Add $- 7 x$ and $x$ .

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

Step 3

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

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

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

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

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

Factor $3$ out of $9$ .

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

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

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

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

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

Step 4

Factor.

Factor by grouping.

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 $b = - 2$ .

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

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

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

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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 $x$ .

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

$- x + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Step 7

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

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

$x + 3 = 0$

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.

$x = 1, - 3$

Solve by Factoring 7-7x=(3x+2)(x-1)

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