# long division

• December 15th 2008, 03:47 AM
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 .
• December 15th 2008, 04:21 AM
Moo
Quote:

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.
• December 15th 2008, 04:33 AM
(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 $ax^3 - 8x^2 + bx + 6 = (x^2 - 2x - 3)(3x - 2) \, ....$