Tony and Mike, factored the trinomial
![8x^2 - 12x - 8](https://img.qammunity.org/2019/formulas/mathematics/high-school/nzvn639zfqanakplp3o5mtdrewp2h0f6qr.png)
Tony factored it as 4(x - 2)(2x + 1) and
Mike factored it as (x - 2)(8x + 4)
![8x^2 - 12x - 8](https://img.qammunity.org/2019/formulas/mathematics/high-school/nzvn639zfqanakplp3o5mtdrewp2h0f6qr.png)
GCF is 4. We factor out 4
![4(2x^2 - 3x - 2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/p0i2llu6eq0advpcb3l1kr5x4pg8tccqnv.png)
2*-2=-4. We find out two factors whose product is -4 and sum is -3
two factors are -4 and 1. Split middle term -3x using two factors
![4(2x^2 - 4x + 1x - 2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/opz9mxehjl1u2ei2ggymv6yrc5kc4y0kvy.png)
Group first two terms and last two terms
![4[(2x^2 - 4x) + (1x - 2)]](https://img.qammunity.org/2019/formulas/mathematics/high-school/qfap0cdpxhndb3vu1qrcu4gqgl5i5l21vb.png)
Factor out GCF from each group
![4[2x(x - 2) + 1(x - 2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/ftevdqmnyfgtc47kae3hnm95exioqyqby9.png)
4(2x+1)(x-2)
Tony factored it correctly
Mike factored it as (x − 2)(8x + 4)
Mike factor 8x+4 further. GCF of 8 and 4 is 4
So it becomes 4(2x+1)
Mike not factored it completely