How do you divide $\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1}$ $(4x^3 + 2x^2 + 9x)/(x^2+2x+1)$ ?

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How do you divide $\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1}$ $(4x^3 + 2x^2 + 9x)/(x^2+2x+1)$ ?

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Answer 1 · 2016-04-01T21:22:30

$\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1} = 4 x - 6 + \frac{17 x + 6}{x^{2} + 2 x + 1}$ $(4x^3+2x^2+9x)/(x^2+2x+1)=4x-6 + (17x+6)/(x^2+2x+1)$

Explanation:

It is possible to long divide the polynomials or simply their coefficients, but let's just pick out the quotient and remainder step by step:

$\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1}$ $(4x^3+2x^2+9x)/(x^2+2x+1)$

$= \frac{4 x^{3} + 8 x^{2} + 4 x - 6 x^{2} + 5 x}{x^{2} + 2 x + 1}$ $=(4x^3+8x^2+4x-6x^2+5x)/(x^2+2x+1)$

$= \frac{4 x (x^{2} + 2 x + 1) - 6 x^{2} + 5 x}{x^{2} + 2 x + 1}$ $=(4x(x^2+2x+1)-6x^2+5x)/(x^2+2x+1)$

$= 4 x + \frac{- 6 x^{2} - 12 x - 6 + 17 x + 6}{x^{2} + 2 x + 1}$ $=4x+(-6x^2-12x-6+17x+6)/(x^2+2x+1)$

$= 4 x + \frac{- 6 (x^{2} + 2 x + 1) + 17 x + 6}{x^{2} + 2 x + 1}$ $=4x+(-6(x^2+2x+1)+17x+6)/(x^2+2x+1)$

$= 4 x - 6 + \frac{17 x + 6}{x^{2} + 2 x + 1}$ $=4x-6 + (17x+6)/(x^2+2x+1)$

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How do you divide $\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1}$ $(4x^3 + 2x^2 + 9x)/(x^2+2x+1)$ ?

1 Answer

Explanation:

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How do you divide 4x3+2x2+9xx2+2x+1(4x^3 + 2x^2 + 9x)/(x^2+2x+1)?

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How do you divide $\frac{4 x^{3} + 2 x^{2} + 9 x}{x^{2} + 2 x + 1}$ $(4x^3 + 2x^2 + 9x)/(x^2+2x+1)$ ?