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))
(4y - 1)(16y2 + 4y + 1)
2y(2y - 1)(2y - 3)
2y(2y - 1)(2y - 3)
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
(4y + 1)(y + 6)
7(5x + 4y)
(4y + 1)(3y + 2)
(4y - 1)(5y + 2)
(y - 1)(3y + 7)
-4x + 4y = 1
2y + 1
x=4y+1 x=4y-1 No,they have different solutions.