u^2 + uv + v has no simple factorisation.
As a quadratic in u, it can be "factorised" by using the quadratic formula to find the root values (r1 and r2) for u, and then the factorisation would be (u - r1)(u - r2); however the values of the roots are:
r1 = (-v + √(v^2 - 4v))/2
r2 = (-v - √(v^2 - 4v))/2
which leads to the complicated "factorisation":
u^2 + uv + v = (u + (v + √(v^2 - 4v))/2)(u + (v - √(v^2 - 4v))/2)
(c + d)(c + d)
(2x + 3)(x + 1)
(3x + 2)(x + 5)
It can't be factored because its discriminant is less than zero.
5x^3(3x + 2)
6x^2(2x + 5)
10(2p + 3)(p + 4)
Find the GCF. The largest factor that those three terms have in common is 5m3n. Take that out. 5m3n(4m4n2 - 2m + n)
(x + 3)(5x + 2)
it is prime. it cannot factor out.
No.
It is: 1(x4+4y8) and can't be factored any further