How do you divide $\frac{x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}{x^{2} - 3}$ $(x^4-2x^3+2x^2-3x+3)/(x^2-3)$ ?

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How do you divide $\frac{x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}{x^{2} - 3}$ $(x^4-2x^3+2x^2-3x+3)/(x^2-3)$ ?

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Answer 1 · 2017-05-18T20:33:44

Given: $\frac{x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}{x^{2} - 3}$ $(x^4-2x^3+2x^2-3x+3)/(x^2-3)$

Write the divisor with a 0 coefficient for missing term

$\frac{x^{2} + 0 x - 3}{x^{2} + 0 x - 3} \frac{(x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3)}{∣ x x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}$ $color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))color(white)((x^4-2x^3+2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$

To determine the first term in the quotient divide the first term in the dividend by the first term in the divisor $\frac{x^{4}}{x^{2}} = x^{2}$ $x^4/x^2=x^2$ write $x^{2}$ $x^2$ in the quotient:

$\frac{x^{2} + 0 x - 3}{x^{2} + 0 x - 3} \frac{x^{2} 2 x^{3} + 2 x^{2} - 3 x + 3}{∣ x x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}$ $color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2color(white)(2x^3+2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$

Multiply the term in the quotient by the divisor:

$x^{2} (x^{2} + 0 x - 3) = x^{4} + 0 x^{3} - 3 x^{2}$ $x^2(x^2+0x-3) =x^4+0x^3-3x^2$

Make it negative:

$- x^{4} - 0 x^{3} + 3 x^{2}$ $-x^4-0x^3+3x^2$

Write it below the dividend:

$\frac{x^{2} + 0 x - 3}{x^{2} + 0 x - 3} \frac{x^{2} 2 x^{3} + 2 x^{2} - 3 x + 3}{∣ x x^{4} - 2 x^{3} + 2 x^{2} - 3 x + 3}$ $color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2color(white)(2x^3+2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$

Perform the addition and bring down the x term:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2color(white)(2x^3+2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$

To find the next term in the quotient, divide the first term in the sum by the first term of the divisor:

$(-2x^3)/x^2=-2x$

Write it as the next term of the quotient:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2xcolor(white)(2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$

Multiply the term in the quotient by the divisor:

$-2x(x^2+0x-3) =-2x^3+0x^2+6x$

Make it negative:

$2x^3+0x^2-6x$

Write it below the sum:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2xcolor(white)(2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$
$color(white)(...............................)ul(2x^3+0x^2-6x)$

Perform the additions and bring down the constant term:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2xcolor(white)(2x^2-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$
$color(white)(...............................)ul(2x^3+0x^2-6x)$
$color(white)(..........................................)5x^2-9x+3$

To find the next term in the quotient, divide the first term in the sum by the first term of the divisor:

$(5x^2)/x^2=5$

Write it as the next term of the quotient:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2x+5color(white)(-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$
$color(white)(...............................)ul(2x^3+0x^2-6x)$
$color(white)(..........................................)5x^2-9x+3$

Multiply the term in the quotient by the divisor:

$5(x^2+0x-3)=5x^2+0x-15$

Make it negative:

$-5x^2+0x+15$

Write it below the sum:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2x+5color(white)(-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$
$color(white)(...............................)ul(2x^3+0x^2-6x)$
$color(white)(..........................................)5x^2-9x+3$
$color(white)(.......................................)ul(-5x^2+0x+15)$

Perform the addition:

$color(white)( (x^2+0x-3)/color(black)(x^2+0x-3))(x^2-2x+5color(white)(-3x+3))/(| color(white)(x)x^4-2x^3+2x^2-3x+3)$
$color(white)(....................)ul(-x^4-0x^3+3x^2)$
$color(white)(..........................)-2x^3+5x^2-3x$
$color(white)(...............................)ul(2x^3+0x^2-6x)$
$color(white)(..........................................)5x^2-9x+3$
$color(white)(.......................................)ul(-5x^2+0x+15)$
$color(white)(...............................................)-9x+18$

Because $-9x+18$ is of order 1 and the divisor is of order 2, we stop and declare $-9x+18$ to be the remainder.

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