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

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Answer 1 · 2018-07-25T13:46:30

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

Perform a long division

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

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

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

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

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

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

$(x^4+0x^3-2x^2+4x+1)/(x^3-x^2-x+1)=x+1+(4x)/(x^3-x^2-x+1)$

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