How do you divide (x ^ 10 + x ^ 8) / (x - 1)?

1 Answer
May 31, 2016

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

Explanation:

Given p_n(x) and (x-x_0) then can be written

p_n(x)=(x-x_0)p_{n-1}(x) + p_0

In this case we have

p_{10}(x) = x^{10}+x^8
x_0 = 1
p_9(x) = sum_{i = 0}^9 a_i x^i

equating the coefficients

p_{10}(x) = x^{10}+x^8 = (x-1)( sum_{i = 0}^9 a_i x^i) + p_0
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