Given p_n(x)pn(x) and (x-x_0)(x−x0) then can be written
p_n(x)=(x-x_0)p_{n-1}(x) + p_0pn(x)=(x−x0)pn−1(x)+p0
In this case we have
p_{10}(x) = x^{10}+x^8p10(x)=x10+x8
x_0 = 1x0=1
p_9(x) = sum_{i = 0}^9 a_i x^ip9(x)=9∑i=0aixi
equating the coefficients
p_{10}(x) = x^{10}+x^8 = (x-1)( sum_{i = 0}^9 a_i x^i) + p_0p10(x)=x10+x8=(x−1)(9∑i=0aixi)+p0
we get
{(a_0 - p_0=0), (-a_0 + a_1=0), (-a_1 + a_2=0),( -a_2 + a_3=0), (-a_3 + a_4=0),( -a_4 + a_5=0), (-a_5 + a_6=0), (-a_6 + a_7=0), (1 - a_7 + a_8=0), (-a_8 + a_9=0), (1 - a_9=0):}
Solving for a_i, p_0 we get
{a_9= 1, a_8 = 1, a_7 = 2, a_6 = 2, a_5= 2, a_4= 2, a_3= 2,
a_2 =2, a_1 =2, a_0= 2, p_0= 2}
quotient
p_9(x) = x^9+x^8+2x^7+2 x^6+2 x^5+2 x^4+2 x^3+2x^2+2 x + 2
and remainder
p_0 = 2
