Using the defining relationship Pn+2(x)=xPn+1(x)−Pn(x) we get the sequence:
P0
= 0
P1
= 1
P2
= x
P3
= x2 − 1
P4
= x3 − 2x
P5
= x4 − 3x2 +1
P6
= x5 − 4x3 +3x  …

Using the defining relationship we can express P6 in terms of previous polynomials in the sequence.
P6
= xP5−P4
= (x2−1)P4−xP3
= P3P4 − P2P3
= P3(P4−P2) .
This shows that P3 is a factor of P6 so all the roots of P3 are roots of P6.

Similarly we can express P8 in terms of previous polynomials in the sequence.
P8
= xP7−P6
= (x2−1)P6−xP5
= (x3−2x)P5−(x2−1)P4
= P4(P5−P3) .
This shows that P4 is a factor of P8 so all the roots of P4 are roots of P8.

Again we can express P10 in terms of previous polynomials in the sequence.
P10
= xP9−P8
= (x2−1)P8−xP7
= (x3−2x)P7−(x2−1)P6
= (x4 − 3x2 +1)P6 − (x3 − 2x)P5
= P5P6−P4P5
= P5(P6−P4).
This shows that P5 is a factor of P10 so all the roots of P5 are roots of P10.

This suggests a conjecture that P2k=Pk(Pk+1−Pk−1) where k is any natural number. This is true but the proof is beyond the scope of school mathematics.