Question

Factorize: 4z3-4z2-11z+6

Answer 1

let f (z)=4z3-4z2=11z+6
put z=2 in f (z),we get
f (2)=4 (2)3-4 (2)2-11 (2)+6
=4(8)-4 (4)-22+6
=32-16-22+6=0
hence,(z-2) Is a factor of (z)
now let's divide f (z) with (z-2)
f (z)=(z-2)(4z2+4z-3)
=(z-2)(4z2+6z-2z-3)
=(z-2)[2z (z+3)-1 (2z+3)]
(z-2)(z+3)(2z-1)

Answer 2

@

f (z)=4z³-4z²-11z+6
=> Putting z=2
=> f (2)=4 (2)3-4 (2)2-11 (2)+6
=> 4(8)-4 (4)-22+6
=> 32-16-22+6=0
=> (z-2) Is a factor of (z)
=> now let's divide f (z) with (z-2)
We get
f (z)=(z-2)(4z2+4z-3)
=> (z-2)(4z2+6z-2z-3)
=> (z-2)[2z (z+3)-1 (2z+3)]
=> (z-2)(z+3)(2z-1) Ans

@:-)