How do you divide $\frac{x^{3} + x^{2} + 4 x - 6}{x^{2} - 3 x + 2}$ $(x^3 + x^2 +4x -6) / (x^2 -3x +2)$ using polynomial long division?

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Answer 1 · 2016-12-23T04:02:31

The quotient is $= (x + 4)$ $=(x+4)$ and the remainder $= 14 (x - 1)$ $=14(x-1)$

Let's do the long division

$a a a a$ $color(white)(aaaa)$ $x^{3} + x^{2} + 4 x - 6$ $x^3+x^2+4x-6$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $x^{2} - 3 x + 2$ $x^2-3x+2$

$a a a a$ $color(white)(aaaa)$ $x^{3} - 3 x^{2} + 2 x$ $x^3-3x^2+2x$ $a a a a a a$ $color(white)(aaaaaa)$ $∣$ $∣$ $x + 4$ $x+4$

$a a a a a$ $color(white)(aaaaa)$ $0 + 4 x^{2} + 2 x - 6$ $0+4x^2+2x-6$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 4 x^{2} - 12 x + 8$ $+4x^2-12x+8$

$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $0 + 14 x - 14$ $0+14x-14$

Therefore,

$\frac{x^{3} + x^{2} + 4 x - 6}{x^{2} - 3 x + 2} = x + 4 + \frac{14 (x - 1)}{x^{2} - 3 x + 2}$ $(x^3+x^2+4x-6)/(x^2-3x+2)=x+4+(14(x-1))/(x^2-3x+2)$

$= x + 4 + \frac{14 (x - 1)}{(x - 2) (x - 1)}$ $=x+4+(14(x-1))/((x-2)(x-1))$

The quotient is $= (x + 4)$ $=(x+4)$ and the remainder $= 14 (x - 1)$ $=14(x-1)$

How do you divide x3+x2+4x−6x2−3x+2(x^3 + x^2 +4x -6) / (x^2 -3x +2) using polynomial long division?