You have \(\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4\).
Do you see that there is a "(2x+ 4)" in both terms? One is to the 6th power and the other is to the 7th power. \(\displaystyle (2x+ 4)^7= (2x+ 4)^6(2x+ 4)\) so there is \(\displaystyle (2x+ 4)^6\) in both terms and we can factor that out, leaving 2x+ 4 to the first power in the second term.
Do you see that there is a "(3x- 2)" in both terms? One is to the 5th power and the other is to the 4th power. \(\displaystyle (3x- 2)^5= (3x- 2)^4(3x- 3)\) so there is \(\displaystyle (3x- 2)^4\) in both terms so we can factor that out, leaving 3x- 2 to the first power in the first term.
\(\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4[(14(3x- 2)+ 15(2x+ 4)]\).
Now we need to "combine" what was left:
14(3x- 2)+ 15(2x+ 4)= 42x- 28+ 30x+ 60= 72x+ 32 so
\(\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= (2x+ 4)^6(3x- 2)^4(72x+ 32)\).
You could, if you like, also take common factors out of the coefficients. 2x+ 4= 2(x+ 2) so \(\displaystyle (2x+ 4)^6= 2^6(x+ 2)^6= 64(x+ 2)^5\). 72= 8(9) and 32= 4(8) so 72x+ 32= 8(9x+ 4). 64(8)= 512.
\(\displaystyle 14(2x+ 4)^6(3x- 2)^5+ 15(2x+ 4)^7(3x- 2)^4= 512(x+ 2)^6(3x- 2)^4(9x+ 4)\).