2x3 - 5x2 + 5x - 2
= x(2x2-5x+3) + 2x - 2
= x(2x-3)(x-1) + 2(x-1)
= (x-1)[x(2x-3)+2]
= (x-1)(2x2-3x+2)
I did it by trial and error, through experience -- no set procedure. (2x2-3x+2) cannot be factored, which is evidenced by calculating (b2-4ac), where a = 2; b=-3; and c=2 from the expression (2x2-3x+2). (b2-4ac) = (9-16) = -7, which has no real square root value. That is, sqrt(-7) is an imaginary number.
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Factor out the GCF and get X(X2-X+1).
n(2n - 1)(2n + 7)
y^3 + y^2 - 6 (y+2) (y-3) i think
6(b - ac + b2 - bc)
(x - 1)(5x + 7)
(a - 2)(a^2 + 6)
Factor out the GCF and get X(X2-X+1).
x(x - 19)(x - 1)
(x - 2)(x^2 + x + 3)
(x + 2)(3x - 1)(3x + 1)
(x - 5)(x^2 + 1)
n(2n - 1)(2n + 7)
You have to put your heart into it!
x³ + x² - 3x - 3 = (x + 1)(x + √3) (x - √3).
x³ - x² + x - 3 (x² + 1)(x - 1) - 2
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
y^3 + y^2 - 6 (y+2) (y-3) i think