16y^2(8y)=0? Factor out an 8y to give you 8y(2y(2))=0 Use order of operations to give you 8y(4y)=0 Distribute to give you 32y=0 and ultimately the answer will be 0 If there is a missing plus or minus operator between 16y2 and the 8y, then it can be factorised by taking out common factors: 16y2 ± 8y = 0 ⇒ 8y(2y ± 1) = 0
2(8y + 1)
If: -8y-7 = -8+8y Then: -7+8 = 8y+8y And: 16y = 1 So: y = 1/16
8y squared + 52y - 180
It cannot be done using real numbers. However, it can be done using complex numbers: 16y² + 4y + 1 = ¼(8y + (1 + i√3))(8y + (1 - i√3))
x + 8y = 3 therefore 2x + 16y = 6 (a) 2x + 3y = -7 (b) Subtract (b) from (a): 13y = 13 so y = 1 and x = -5
0
2y(y + 3)(y - 7)
7
LCM(10y2 , 8y) = 40y2
I feel like there is something wrong with your question because subtracting y2 on both sides leaves you with 8y=0; then, y=0
The only thing you can do to this is factor it... (4x + 8y)(4x - 8y)