How do you long divide $(x^3+3x^2+x)div (x+2)$ ?

Question

1536 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-08T16:17:11

The quotient is $=x^2+x-1$ and the remainder is $=2$

You can either do a long division or use the remainder theorem

Let's do the long division

$color(white)(aaaa)$ $x^3+3x^2+x$ $color(white)(aaaa)$ $∣$ $x+2$

$color(white)(aaaa)$ $x^3+2x^2$ $color(white)(aaaaaaaa)$ $∣$ $x^2+x-1$

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

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

$color(white)(aaaaaaaaa)$ $+0-x$

$color(white)(aaaaaaaaaaaaa)$ $-x$

$color(white)(aaaaaaaaaaaaa)$ $-x-2$

$color(white)(aaaaaaaaaaaaaaa)$ $0+2$

So

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

Let's try the remainder theorem

Let $f(x)=x^3+3x^2+x$

Then. $f(-2)=(-2)^3+3*(-2)^2-2$

$=-8+12-2=2$

The remainder is $=2$

How do you long divide (x^3+3x^2+x)div (x+2)(x^3+3x^2+x)div (x+2)?