2y2 + 9y - 2 this will not factor using integers what is what most algebra teachers will tell you is 'prime'
but ... make an equation ... 2y2 + 9y - 2 = 0
use the quadratic formula to solve
y = [-9 (+/-) sq rt (81 -4x2(-2))] / (2x2)
y = [-9 (+/-) sq rt (97)] / (4) now separate these
y = [-9 + sq rt (97)] / (4)
y = [-9 - sq rt (97)] / (4) rewrite as equal to zero
(y - [-9 + sq rt (97)] / (4)) =0
(y - [-9 - sq rt (97)] / (4)) = 0 apply zero product property in reverse
(y - [-9 + sq rt (97)] / (4))(y - [-9 - sq rt (97)] / (4)) = 0
these, as weird as they may look, are the factors.
(5x - 2y)(5x + 2y)
If that's -18y2, the answer is (3x - 2y)(x + 9y)
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)
12y3 - 30y2 + 12yFactor out a y:y(12y2 - 30y + 12)Factor out a 6:6y(2y2 -5y + 2)Factor 2y2 - 5y + 26y(2y - 1)(y - 2)
It is: 3y(2y-3x) when factored
(2y - 5)(y - 2)
(5x - 2y)(5x + 2y)
2y + 1
2y^2+5y+2=(2y+1)(y+2)
6xyz(3x + 2y + z)
y^(3) - 2y^(2) + 4y Factor out 'y' Hence y(y^2 - 2y + 4) This will not factor any further.
It is the difference of two squares which is: (x+2y)(x-2y)
If that's -18y2, the answer is (3x - 2y)(x + 9y)
(2y+1)(y−5)
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)
Pull out the common factor of 2 ... 2(y^2+6y-16) Then just factor (y^2+6y-16) ... (y-2)(y+8)
11y(y - 1)(2y^2 + 2y - 1)