Question

Verify that 3, -2, 1 are the zeroes of the cubic polynomial $p\left(x\right)=x^3-2x^2-5x+6$

Answer 1

p(x)=$p\left(x\right)=x^3-2x^2-5x+6$

At x=3
P(3)=$(3{)}^3-2(3{)}^2-5\left(3\right)+6$
=27-18-15+6
=0
since p(3)=0
hence 3 is a zero of p(x)

at x=-2
p(-2)=$(-2{)}^3-2(-2{)}^2-5(-2)+6$
=-8-8+10+6
=0
since p(-2)=0
hence -2 is a zero of p(x)

at x=1
p(1)= $(1{)}^3-2(1{)}^2-5\left(1\right)+6$
=1-2-5+6
=0
Since p(1)=0
hence 1 is a zero of p(x)