Question

p (y)= y^2+43y+22 find the zeros​

Answer 1

p(y)=y^2+43y+22

we first equate it to 0

y^2+43y+22=0

=y^2+43y=0-22=-22

=y^2+y=-22/43

=2y =-22/43 ×1/y

=2y^2=-22/43

=y^2=-22/43 ×2=-44/43

so threfore the zero of the polynomyal is square root of -44/23