If that were z^2 - 16z + 64, it would factor very neatly to (z - 8)(z - 8) or (z - 8)^2
As it is, you could use z(z^2 - 16) + 64 or the much more cumbersome
-1/27 (3 z+2 2^(2/3) (3 (9-sqrt(69)))^(1/3)+4 3^(2/3) (2/(9-sqrt(69)))^(1/3)) (-9 z^2+(12 3^(2/3) (2/(9-sqrt(69)))^(1/3)+6 2^(2/3) (3 (9-sqrt(69)))^(1/3)) z-8 2^(1/3) (3 (9-sqrt(69)))^(2/3)-48 3^(1/3) (2/(9-sqrt(69)))^(2/3)+48)
ax + ay - AZ = a(x + y - z)
Their product.
2 x 2 x 2 x 3 x z x z = 24z2
To raise to a power it's a lot clearer if you use ^ on a PC. For example, x^2 for x squared. I think you want to factorise 4x^2 - 25(y^2)(z^2). We recognise this as a difference of squares: (2x)^2 - (5yz)^2 and so it factorises as (2x - 5yz)(2x + 5yz).
-3(x + y + z)
This is a difference of two sqaures. 49y2 - z2 = (7y - z)(7y + z)
It can't be factored because its discriminant is less than zero.
(z-9)(z+3
Using the quadratic formula you get z≅4.91547594742265 or z≅-0.91547594742265
6xyz(3x + 2y + z)
(c - z)(c^2 + cz + z^2)
The only common factor to all terms is yz. → xy³z² + y²z + xyz = yz(xy²z + y + x)
z^2
ax + ay - AZ = a(x + y - z)
4z squared (z+4) (z+4)
x squared + y squared = z squared.
(z=-3) and (z=7)