(x-m)2=x2-2mx+m2.
Divide into x4-4x3+hx2-6x+2 using algebraic long division:
x2+(2m-4)x+h+3m2-8m
x2-2mx+m2 ) x4 -4x3 +hx2 -6x +2
x4-2mx3 +m2x2
(2m-4)x3 +(h-m2)x2 -6x
(2m-4)x3-2m(2m-4)x2 +m2(2m-4)x
(h+3m2-8m)x2 +(-2m3+4m2-6)x +2
(h+3m2-8m)x2-2m(h+3m2-8m)x+m2(h+3m2-8m)
0
Because the remainder must be zero:
-2m3+4m2-6+2mh+6m3-16m2=0=4m3-12m2+2mh-6, and 2-m2h-3m4+8m3=0.
We can eliminate h: m(4m3-12m2+2mh-6)+2(2-m2h-3m4+8m3)=0,
4m4-12m3+2m2h-6m+4-2m2h-6m4+16m3=-2m4+4m3-6m+4=0=-m4+2m3-3m+2. m=1 is a root so m-1 is a factor, so after factorising we get:
(m-1)(-m3+m2+m-2)=0. We're told that m is a natural number, so m=1.
Now we can find h by substituting for m:
2-h-3+8=0, h=2-3+8=7.
We have x4-4x3+7x2-6x+2 and (x-1)2 is the factor.
SOLUTION: m=1, h=7.
CHECK
(x-1)(x3-3x2+4x-2)=(x-1)2(x2-2x+2)=x4-4x3+7x2-6x+2