Question

Why is the odd monic polynomial of least degree with a triple root of x=-2 and a single root of x=1 : P(x) =`x(x-1)(x+1)(x-2)^3(x+2)^3` ?

Answer 1

See explanation...

Explanation:

`f(x)` is an odd function if and only if `f(-x) = -f(x)` for all `x` in the domain of `f(x)`.

Note that zeros correspond exactly to linear factors.

So a zero `a` of multiplicity `n` corresponds to a factor `(x-a)^n`

The monic polynomial of least degree with triple zero `x=-2` and single zero `x=1` is:

`(x-1)(x+2)^3`

If it is odd, then it must have a triple zero `x=2` and `x=-1` too, giving us:

`(x-1)(x+1)(x-2)^3(x+2)^3`

Finally, the polynomial's value at `x=0` must be `0` too, since if is was non-zero the function would not be odd. So there is a factor `(x-0) = x` too, giving us:

`x(x-1)(x+1)(x-2)^3(x+2)^3`

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this one.

Note that this polynomial is an odd polynomial, as we find by substituting `-x` for `x` ...

`(-x)((-x)-1)((-x)+1)((-x)-2)^3((-x)+2)^3`

`=(-x)(-(x+1))(-(x-1))(-(x+2))^3(-(x-2))^3`

`=-x(x+1)(x-1)(x+2)^3(x-2)^3`

`=-x(x-1)(x+1)(x-2)^3(x+2)^3`