How do you divide $6 x^{4} + x^{3} + 1 \div 2 x^{2} - 2$ $6x^4+x^3+1 div 2x^2-2$ ?

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Answer 1 · 2015-12-16T04:42:23

$\frac{(x^{2} - x + 1) (6 x + 1)}{2 (x - 1)}$ $((x^2-x+1)(6x+1))/(2(x-1))$

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

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

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

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

$=(color(red)cancelcolor(black)((x+1))(x^2-x+1)(6x+1))/(2color(red)cancelcolor(black)((x+1))(x-1))$

$=((x^2-x+1)(6x+1))/(2(x-1))$

How do you divide 6x^4+x^3+1 div 2x^2-26x4+x3+1÷2x2−26x^4+x^3+1 div 2x^2-2?