1. ## long division

Divide x^4+6x^3-5x^2+13x+1 by X^2+2 using long division

Can someone solve this for me? I tried but my answer doesn't match the answer in the book.

2. Originally Posted by cityismine
Divide x^4+6x^3-5x^2+13x+1 by X^2+2 using long division

Can someone solve this for me? I tried but my answer doesn't match the answer in the book.
I have attached a word document with the solution for you.

3. Thx, that's the same answer I got as well. I guess the answer in the book is wrong.

4. Hello, cityismine!

There are many chances for error with the "spaces" involved.
Let me give it a try . . .

Divide: . $(x^4+6x^3-5x^2+13x+1) \div (x^2+2)$

$\begin{array}{cccccccccccc}
& & & & & & x^2 & + & 6x & - & 7 \\
& & --& --&--&--&--&--&--&--&-- \\
x^2+2 & | & x^4 & + & 6x^3 & - & 5x^2 & + & 13x & + & 1 \\
& & x^4 & & & + & 2x^2 \\
& & --&--&--&--&-- \\
& & & & 6x^3 & - & 7x^2 & + & 13x \\
& & & & 6x^3 & & & + & 12x \\
& & & & --&--&--&--&-- \\
& & & & & - & 7x^2 & + & x & + & 1 \end{array}$

. . . . . . . . . . . . . . . . . $\begin{array}{cccccccccccc}\quad\quad & \quad &\quad &\quad &\quad & - & 7x^2 &\quad &\quad & - & 14\\
& & & & & & --&--&--&--&-- \\
& & & & & & & & x & + & 15\end{array}$

Therefore: . $\frac{x^4+6x^3 - 5x^2 + 13x + 1}{x^2+2} \;\;=\;\;x^2+6x-7 \,+\, \frac{x+15}{x^2+2}$

5. Originally Posted by Soroban
Hello, cityismine!

There are many chances for error with the "spaces" involved.
Let me give it a try . . .

$\begin{array}{cccccccccccc}$ $
& & & & & & x^2 & + & 6x & - & 7 \\
& & --& --&--&--&--&--&--&--&-- \\
x^2+2 & | & x^4 & + & 6x^3 & - & 5x^2 & + & 13x & + & 1 \\
& & x^4 & & & + & 2x^2 \\
& & --&--&--&--&-- \\
& & & & 6x^3 & - & 7x^2 & + & 13x \\
& & & & 6x^3 & & & + & 12x \\
& & & & --&--&--&--&-- \\
& & & & & - & 7x^2 & + & x & + & 1 \end{array}" alt="
& & & & & & x^2 & + & 6x & - & 7 \\
& & --& --&--&--&--&--&--&--&-- \\
x^2+2 & | & x^4 & + & 6x^3 & - & 5x^2 & + & 13x & + & 1 \\
& & x^4 & & & + & 2x^2 \\
& & --&--&--&--&-- \\
& & & & 6x^3 & - & 7x^2 & + & 13x \\
& & & & 6x^3 & & & + & 12x \\
& & & & --&--&--&--&-- \\
& & & & & - & 7x^2 & + & x & + & 1 \end{array}" />

. . . . . . . . . . . . . . . . . $\begin{array}{cccccccccccc}\quad\quad & \quad &\quad &\quad &\quad & - & 7x^2 &\quad &\quad & - & 14\\$ $
& & & & & & --&--&--&--&-- \\
& & & & & & & & x & + & 15\end{array}" alt="
& & & & & & --&--&--&--&-- \\
& & & & & & & & x & + & 15\end{array}" />

Therefore: . $\frac{x^4+6x^3 - 5x^2 + 13x + 1}{x^2+2} \;\;=\;\;x^2+6x-7 \,+\, \frac{x+15}{x^2+2}$
Same as mine