How do you long divide $(x^{4} - 2 x^{2} + 4 x + 1) \div (x^{3} - x^{2} - x + 1)$ $(x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)$ ?

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How do you long divide $(x^{4} - 2 x^{2} + 4 x + 1) \div (x^{3} - x^{2} - x + 1)$ $(x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)$ ?

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Answer 1 · 2018-07-25T13:46:30

The remainder is $= 4 x$ $=4x$ and the quotient is $= x + 1$ $=x+1$

Explanation:

Perform a long division

$a a a a$ $color(white)(aaaa)$ $x^{4} + 0 x^{3} - 2 x^{2} + 4 x + 1$ $x^4+0x^3-2x^2+4x+1$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x^{3} - x^{2} - x + 1$ $x^3-x^2-x+1$

$a a a a$ $color(white)(aaaa)$ $x^{4} - 1 x^{3} - 1 x^{2} + 1 x$ $x^4-1x^3-1x^2+1x$ $a a a a a a a$ $color(white)(aaaaaaa)$ $∣$ $|$ $x + 1$ $x+1$

$a a a a a$ $color(white)(aaaaa)$ $0 + 1 x^{3} - 1 x^{2} + 3 x + 1$ $0+1x^3-1x^2+3x+1$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 1 x^{3} - 1 x^{2} - x + 1$ $+1x^3-1x^2-x+1$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 0 x^{3} - 0 x^{2} + 4 x + 0$ $+0x^3-0x^2+4x+0$

The remainder is $= 4 x$ $=4x$ and the quotient is $= x + 1$ $=x+1$

$\frac{x^{4} + 0 x^{3} - 2 x^{2} + 4 x + 1}{x^{3} - x^{2} - x + 1} = x + 1 + \frac{4 x}{x^{3} - x^{2} - x + 1}$ $(x^4+0x^3-2x^2+4x+1)/(x^3-x^2-x+1)=x+1+(4x)/(x^3-x^2-x+1)$

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How do you long divide $(x^{4} - 2 x^{2} + 4 x + 1) \div (x^{3} - x^{2} - x + 1)$ $(x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)$ ?

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How do you long divide (x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)(x4−2x2+4x+1)÷(x3−x2−x+1)(x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)?

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Explanation:

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How do you long divide $(x^{4} - 2 x^{2} + 4 x + 1) \div (x^{3} - x^{2} - x + 1)$ $(x^4 - 2x^2 + 4x + 1) ÷ (x^3-x^2-x+1)$ ?