How do you find the quotient of $(2w^3+3w^2-5w+1)div(w+3)$ using long division?

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Answer 1 · 2017-02-07T14:21:50

The quotient is $=color(red)(2w^2-3w+4)$

Let's do the long division

$color(white)(aaaa)$ $2w^3+3w^2-5w+1$ $color(white)(aaaa)$ $|$ $color(blue)(w+3)$

$color(white)(aaaa)$ $2w^3+6w^2$ $color(white)(aaaaaaaaaaaa)$ $|$ $color(red)(2w^2-3w+4)$

$color(white)(aaaaaa)$ $0-3w^2-5w$

$color(white)(aaaaaaaa)$ $-3w^2-9w$

$color(white)(aaaaaaaaaa)$ $-0+4w+1$

$color(white)(aaaaaaaaaaaaaa)$ $+4w+12$

$color(white)(aaaaaaaaaaaaaaaa)$ $+0-11$

Therefore

The remainder is $=-11$

The quotient is $=color(red)(2w^2-3w+4)$

How do you find the quotient of (2w^3+3w^2-5w+1)div(w+3)(2w^3+3w^2-5w+1)div(w+3) using long division?