Hey Trefoil2727.
You should try dividing P(x) by (x-m)^2 and note that your remainder must be 0. Once you have done this you will end up with a quadratic which you can find the roots for using the standard result.
The polynomial P(x) =x^{4}-4x^{3}+hx^{2}-6x+2 has a factor in the form (x-m)^{2}, find the values of m and h.
P(x) =x^{4}-4x^{3}+hx^{2}-6x+2
P(m) = 0
m^{4}-4m^{3}+hm^{2}-6m+2=0
P’(x) =4x^{3}-12x^{2}+2hx-6
P’(m)=4m^{3}-12m^{2}+2hm-6=0
-m^{5}+4m^{4}+6m^{2}-2m=-2m^{5}+6m^{4}+3m^{2}
m^{5}-2m^{4}+3m^{2}-2m=0
÷m, m^{4}-2m^{3}+3m-2=0
Try m=1, (1)^{4}-2(1)^{3}+3(1)-2=0
Hence, m=1, h=7
I think there is something wrong in my solution, can anyone help?