Question

4z^2+19z+12 factorize​

Answer 1

Step-by-step explanation:

using discrimination formula

compare it with az^2+bz+c we get a=4,b=19,c=12

z= [-b+-√(b^2-4ac)]/2a

=[ -19+-√((19)^2-4*4*12)]/2*4

=[-19+-√(361-169)]/8

=[-19+-√169]/8

=[-19+-13]8

= (-19+13)/8; (-19-13)/8

= (-6/8); (-32/8)

= (-3/4); (-4)

factor is (x+3/4)(x+4)