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# ?

1 Answer
Dec 25, 2017

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#