Question

4s square -4s+1 find zero of the polynomial​

Answer 1

Step-by-step explanation:

4s

$4 {x}^{2} - 4x - + 1 \\ 4 {x }^{2} - 2x - 2x + 1 \\ 2x(2x - 1) - 1(2x - 1) \\ (2x - 1)(2x - 1) \\ \\ x = 1 \div 2 \: \: \: x = 1 \div 2$