Question

If \times $p\left(x\right)=x^3-3x^2+2x$, find p(0), p(1), p(2). What do you conclude?

Answer 1

$p\left(x\right)=x^3-3x^2+2x$----- (1)

Putting x = 0 in (1), we get

$p\left(0\right)=0^3-3\times 0^2+2\times 0=0$

Thus, x = 0 is a zero of p(x).

Putting x = 1 in (1), we get

$p\left(1\right)=1^3-3\times 1^2+2\times 1=1-3+2=0$

Thus, x = 1 is a zero of p(x).

Putting x = 2 in (1), we get

$p\left(2\right)=2^3-3\times 2^2+2\times 2=8-3\times 4+4=8-12+4=0$

Thus, x = 2 is a zero of p(x).

We can conclude that 0, 1 and 2 are zeroes of the polynomial $p\left(x\right)=x^3-3x^2+2x$