Solve by Factoring (x-2)(x-1)=6

$(x - 2) (x - 1) = 6$

Step 1

Subtract $6$ from both sides of the equation.

$(x - 2) (x - 1) - 6 = 0$

Step 2

Simplify $(x - 2) (x - 1) - 6$ .

Simplify each term.

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

Apply the distributive property.

$x (x - 1) - 2 (x - 1) - 6 = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 2 (x - 1) - 6 = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 2 x - 2 \cdot - 1 - 6 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 2 x - 2 \cdot - 1 - 6 = 0$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x - 2 x - 2 \cdot - 1 - 6 = 0$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - 2 x - 2 \cdot - 1 - 6 = 0$

Multiply $- 2$ by $- 1$ .

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

Subtract $2 x$ from $- x$ .

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

Subtract $6$ from $2$ .

$x^{2} - 3 x - 4 = 0$

Step 3

Factor $x^{2} - 3 x - 4$ using the AC method.

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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

Step 4

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

$x - 4 = 0$

$x + 1 = 0$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve by Factoring (x-2)(x-1)=6

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