How do you divide $( -3x^3+ x^2-3x-28 )/(x + 1 )$ ?

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-21T10:22:59

The quotient is $=-3x^2+4x-7$ and the remainder is $=-21$

Let's perform the long division

$color(white)(aaaa)$ $x+1$ $|$ $-3x^3+x^2-3x-28$ $color(white)(aaaa)$ $|$ $-3x^2+4x-7$

$color(white)(aaaaaaaaaa)$ $-3x^3-3x^2$

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

$color(white)(aaaaaaaaaaaaaaa)$ $+4x^2+4x$

$color(white)(aaaaaaaaaaaaaaaaa)$ $+0-7x-28$

$color(white)(aaaaaaaaaaaaaaaaaaaaa)$ $-7x-7$

$color(white)(aaaaaaaaaaaaaaaaaaaaaa)$ $-0-21$

The quotient is $=-3x^2+4x-7$ and the remainder is $=-21$

How do you divide ( -3x^3+ x^2-3x-28 )/(x + 1 )( -3x^3+ x^2-3x-28 )/(x + 1 )?