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

May 31, 2016

Quotient ${p}_{9} \left(x\right) = {x}^{9} + {x}^{8} + 2 {x}^{7} + 2 {x}^{6} + 2 {x}^{5} + 2 {x}^{4} + 2 {x}^{3} + 2 {x}^{2} + 2 x + 2$ and remainder ${p}_{0} = 2$

#### Explanation:

Given ${p}_{n} \left(x\right)$ and $\left(x - {x}_{0}\right)$ then can be written

${p}_{n} \left(x\right) = \left(x - {x}_{0}\right) {p}_{n - 1} \left(x\right) + {p}_{0}$

In this case we have

${p}_{10} \left(x\right) = {x}^{10} + {x}^{8}$
${x}_{0} = 1$
${p}_{9} \left(x\right) = {\sum}_{i = 0}^{9} {a}_{i} {x}^{i}$

equating the coefficients

${p}_{10} \left(x\right) = {x}^{10} + {x}^{8} = \left(x - 1\right) \left({\sum}_{i = 0}^{9} {a}_{i} {x}^{i}\right) + {p}_{0}$
we get

$\left\{\begin{matrix}{a}_{0} - {p}_{0} = 0 \\ - {a}_{0} + {a}_{1} = 0 \\ - {a}_{1} + {a}_{2} = 0 \\ - {a}_{2} + {a}_{3} = 0 \\ - {a}_{3} + {a}_{4} = 0 \\ - {a}_{4} + {a}_{5} = 0 \\ - {a}_{5} + {a}_{6} = 0 \\ - {a}_{6} + {a}_{7} = 0 \\ 1 - {a}_{7} + {a}_{8} = 0 \\ - {a}_{8} + {a}_{9} = 0 \\ 1 - {a}_{9} = 0\end{matrix}\right.$

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

${p}_{9} \left(x\right) = {x}^{9} + {x}^{8} + 2 {x}^{7} + 2 {x}^{6} + 2 {x}^{5} + 2 {x}^{4} + 2 {x}^{3} + 2 {x}^{2} + 2 x + 2$

and remainder

${p}_{0} = 2$