Factoring by Grouping

$$\left(x+4\right)\left(x+2\right)$$  

$$\left(x-4\right)\left(x^2-4\right)$$  

$$\left(x+4\right)\left(x^2-4\right)$$  

$$\left(x-4\right)\left(x-2\right)$$  

$$\left(2p^2+3\right)\left(p+2\right)$$  

$$\left(2p^2+6\right)\left(p+4\right)$$  

$$\left(4p^2+3\right)\left(p+2\right)$$  

$$\left(4p^2+8p\right)\left(3p+6\right)$$  

$$\left(x^2-2\right)\left(x-7\right)$$  

$$\left(x^2+2\right)\left(x+7\right)$$  

$$\left(x^2+2\right)\left(x-7\right)$$  

$$\left(x^2-2\right)\left(x+7\right)$$  

$$\left(5n^2+3\right)\left(n-2\right)$$  

$$\left(5n^2-3\right)\left(n-2\right)$$  

$$\left(5n^3+3\right)\left(n-2\right)$$  

$$10n\left(n^2-6\right)$$  

$$\left(x-2\right)$$  

$$\left(x+2\right)$$  

$$\left(x^2+5\right)$$  

$$\left(x^2-5\right)$$  

$$\left(x^2-5\right)\left(x-5\right)$$  

$$\left(x^2+5\right)\left(x-5\right)$$  

$$\left(x^2-5\right)\left(x+5\right)$$  

$$\left(x^2+5\right)\left(x+5\right)$$  

$$\left(5p^2-1\right)\left(4p-5\right)$$  

$$\left(5p^2+1\right)\left(4p+5\right)$$  

$$\left(5p^2+1\right)\left(4p-5\right)$$  

$$\left(5p^2-1\right)\left(4p+5\right)$$  

Yes, it is a factor.

No, it is not a factor.

(3x² + 1) (4x - 3)

(3x² - 1) (4x + 3)

(4x² + 1) (3x - 3)

(4x² - 1) (3x + 3)

(x² + 3) (2x + 5)

(x² - 3) (2x - 5)

(2x² - 3) (x - 5)

(2x² + 3) (x + 5)