Question

find the zeros 2n^2+n-406

Answer 1

=2n²+n-406

=2n²+29n-28n-406

=n(2n+29)-14(2n+29)

=(2n+29)(n-14)

2n+29=0

2n= -29

n= -29/2

n-14=0

n=14

So the zeroes of the polynomial are -29/2 and 14