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Math Help - long division

  1. #1
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    long division

    Find , using long division , the quotient and the remainder of the following :

    (1) \frac{(x^4-2x^3+6x-5)}{(x^2-x-1)}

    Can someone pls show me the working . Sorry , i just couldn't find a way to post my working .

    (2) The expression ax^3-8x^2+bx+6 is divisible by x^2-2x-3 . Find the values of A and B .
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    Find , using long division , the quotient and the remainder of the following :

    (1) \frac{(x^4-2x^3+6x-5)}{(x^2-x-1)}

    Can someone pls show me the working . Sorry , i just couldn't find a way to post my working .
    Take the leading term : x^4
    x^4=x^2 \cdot {\color{red}x^2}

    So write {\color{red}x^2}(x^2-x-1)=x^4-x^3-x^2 \implies x^4=x^2(x^2-x-1)+x^3+x^2 :
    x^4-2x^3+6x-5=[x^2(x^2-x-1)+x^3+x^2]-2x^3+6x-5
    =x^2(x^2-x-1)-x^3+x^2+6x-5

    again, take the leading term of what's left : -x^3
    -x^3=x^2 \cdot \left({\color{red}-x}\right)

    So write {\color{red}-x}(x^2-x-1)=-x^3+x^2+x \implies -x^3=-x(x^2-x-1)-x^2-x :
    x^2(x^2-x-1)-x^3+x^2+6x-5=x^2(x^2-x-1)+[-x(x^2-x-1)-x^2-x]+x^2+6x-5
    =x^2(x^2-x-1)-x(x^2-x-1)+5x-5
    =(x^2-x)(x^2-x-1)+5x-5

    so the division by x^2-x-1 will give a remainder of 5x-5.
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    Re :

    got it , thanks . How about the second one ?
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    [snip]
    (2) The expression ax^3-8x^2+bx+6 is divisible by x^2-2x-3 . Find the values of A and B .
    If you think about it, you'll see that ax^3 - 8x^2 + bx + 6 = (x^2 - 2x - 3)(3x - 2) \, ....
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