Algebra 1 Factoring

$$\left(x+12\right)\left(x+2\right)$$  

$$\left(x+7\right)\left(x+7\right)$$  

$$\left(x+6\right)\left(x+4\right)$$  

$$\left(x+8\right)\left(x+3\right)$$  

$$\left(x+10\right)\left(x+4\right)$$  

$$\left(x-6\right)\left(x-2\right)$$  

$$\left(x+5\right)\left(x+3\right)$$  

$$\left(x+3\right)\left(x-5\right)$$  

$$\left(x+12\right)\left(x+3\right)$$  

$$\left(x-3\right)\left(x-5\right)$$  

$$\left(x^2+2\right)\left(x+8\right)$$  

$$\left(x+2\right)\left(x+8\right)$$  

$$\left(x^2+8\right)\left(x+2\right)$$  

$$\left(2x^2+8\right)\left(x+2\right)$$  

$$\left(x^2+2\right)\left(x+2\right)$$  

$$\left(x+12\right)\left(x+1\right)$$  

$$\left(2x-1\right)\left(x+6\right)$$  

$$\left(2x\right)\left(x+6\right)$$  

$$\left(2x+1\right)\left(x+6\right)$$  

$$\left(2x^2+1\right)\left(x+6\right)$$  

A

B

C

D

(x - 2)(x + 1)

(x - 2)(x - 1)

(x + 2)(x + 1)

(x + 2)(x - 1)

A

B

C

D

In Step 1 because the expression inside the parentheses needs to have a sum of -11x, not 11x.

In Step 2 because the sign on x should be negative and the sign on 2 should be positive.

In Step 3 because there is no need to multiply (x-2) by 1.

In Step 4 because (x-2) should be written first.

A

B

C

D

x - 1

3x - 2

3x + 1

3

x + 1