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  1. #1
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    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.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by cityismine View Post
    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.
    Attached Files Attached Files
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  3. #3
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    Thx, that's the same answer I got as well. I guess the answer in the book is wrong.
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  4. #4
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    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}<br />
& & & & & & x^2 & + & 6x & - & 7 \\<br />
& & --& --&--&--&--&--&--&--&-- \\<br />
x^2+2 & | & x^4 & + & 6x^3 & - & 5x^2 & + & 13x & + & 1 \\<br />
& & x^4 & & & + & 2x^2 \\<br />
& & --&--&--&--&-- \\<br />
& & & & 6x^3 & - & 7x^2 & + & 13x \\<br />
& & & & 6x^3 & & & + & 12x \\<br />
& & & & --&--&--&--&-- \\<br />
& & & & & - & 7x^2 & + & x & + & 1  \end{array}

    . . . . . . . . . . . . . . . . . \begin{array}{cccccccccccc}\quad\quad & \quad &\quad &\quad &\quad & - & 7x^2 &\quad &\quad & - & 14\\<br />
& & & & & & --&--&--&--&-- \\<br />
& & & & & & & & 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}

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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post
    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
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