How do you long divide $\frac{x^{3} + 3 x^{2} + 3 x + 1}{x + 1}$ $(x^3+3x^2+3x+1)/(x+1)$ ?

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How do you long divide $\frac{x^{3} + 3 x^{2} + 3 x + 1}{x + 1}$ $(x^3+3x^2+3x+1)/(x+1)$ ?

2 Answers

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Answer 1 · 2016-11-03T14:23:58

ans. ${(x + 1)}^{2}$ $(x+1)^2$ .

Explanation:

we know ${(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ $(a+b)^3=a^3+3a^2b+3ab^2+b^3$
hence ${(x + 1)}^{3} = x^{3} + 3 x^{2} + 3 x + 1$ $(x+1)^3=x^3+3x^2+3x+1$
so $\frac{x^{3} + 3 x^{2} + 3 x + 1}{x + 1}$ $(x^3+3x^2+3x+1)/(x+1)$
= $\frac{{(x + 1)}^{3}}{x + 1}$ $(x+1)^3/(x+1)$
= ${(x + 1)}^{2}$ $(x+1)^2$ .

Answer 2 · 2016-11-03T17:56:14

$x^{2} + 2 x + 1$ $x^2+2x+1$

Explanation:

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How do you long divide $\frac{x^{3} + 3 x^{2} + 3 x + 1}{x + 1}$ $(x^3+3x^2+3x+1)/(x+1)$ ?

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How do you long divide $\frac{x^{3} + 3 x^{2} + 3 x + 1}{x + 1}$ $(x^3+3x^2+3x+1)/(x+1)$ ?