Grouping by Factoring

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

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

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

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

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

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

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

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

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

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

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

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

$$(x^2-1)(x+2)$$

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

$$(x + 1)(x^2 + 2)$$

$$(x - 1)(x^2 + 2)$$

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

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

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

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

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

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

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

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

Yes, it is a factor.

No, it is not a factor.