Question

If p(x)= x³-3x²+2x, find p(0), p(1), p(2). What do you conclude?

Answer 1

Step-by-step explanation:

p(x)=x^3-3x^2+2x

p(0) =0^3-3×0^2+2×0

=0-0+0

=0

p(1)=1^3-3×1^2+2×1

=1-3×1+2

=1-3+2

=3-3

=0

p(2)=2^3-3×2^2+2×2

=8-12+4

=12-12

=0

Answer 2

Answer:

Step-by-step explanation:

p(x)=x^3-3x^2+2x

p(0) =0^3-3×0^2+2×0

=0-0+0

=0

p(1)=1^3-3×1^2+2×1

=1-3×1+2

=1-3+2

=3-3

=0

p(2)=2^3-3×2^2+2×2

=8-12+4

=12-12

=0

therefore they are zero polynomials