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#