How do you divide $(3x^3+4x^2-2x+5)/(x-1)$ ?

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Answer 1 · 2017-03-14T08:12:59

The remainder is $=10$ and the quotient is $=3x^2+7x+5$

We can perform a long division

$color(white)(aaaa)$ $3x^3+4x^2-2x+5$ $color(white)(aaaa)$ $|$ $x-1$

$color(white)(aaaa)$ $3x^3-3x^2$ $color(white)(aaaaaaaaaaaa)$ $|$ $3x^2+7x+5$

$color(white)(aaaaaa)$ $0+7x^2-2x$

$color(white)(aaaaaaaa)$ $+7x^2-7x$

$color(white)(aaaaaaaaaa)$ $+0+5x+5$

$color(white)(aaaaaaaaaaaaaa)$ $+5x-5$

$color(white)(aaaaaaaaaaaaaaa)$ $+0+10$

The remainder is $=10$ and the quotient is $=3x^2+7x+5$

$(3x^3+4x^2-2x+5)/(x-1)=3x^2+7x+5+10/(x-1)$

How do you divide (3x^3+4x^2-2x+5)/(x-1) (3x^3+4x^2-2x+5)/(x-1) ?