2(5x + 3y)(2x + 5y)
(20x + 3)(1x + 1)
(2x+1)(2x+3) when factored
20x2 + 22xy + 6y2 = 20x2 + 10xy + 12xy + 6y2 = 10x(2x + y) + 6y(2x + y) = (2x + y)(10x + 6y) = 2(2x + y)(5x + 3y)
20x2 - 3x + 10 does not have any real factors.
20X2 + 3010(2X2 + 3)============That's as much factoring as can be done here.
This polynomial doesn't factor. The only thing you can do is take out parts of some terms, e.g. 2(2x3 + 10x2 + x) - 3.
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)
Divide by x: x(x2 - 20x + 19); = x(x -1)(x - 19)
20x2 = 41x+9 20x2-41x-9 = 0 (5x+1)(4x-9) = 0 Therefore: x = -1/5 or x = 9/4
-(5x - 2)(4x - 3)
2x + 6 = 2(x + 3) so 2x + 6 = 0 implies x = -3 Now, substitute x = -3 in 8x3 + 20x2 + 36 which is -8*27 + 20*9 + 36 = -216 + 180 + 36 = 0 So -3 is a root of 8x3 + 20x2 + 36 ie, by the remainder theorem, (x + 3) is a factor. Also, each of the coeffivcients in the expression is even, so 2 is also a factor. So 2*(x + 3) or (2x + 6) is a factor.
10x2-36x+18 Divide all terms by 2: 5x2-18x+9 = (5x-3)(x-3) when factored