How do you divide $(5x^4-3x^3+2x^2-1)div(x^2+4)$ using long division?

Question

5963 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-03T08:01:49

The answer is $=(5x^2-3x-18)+(12x+71)/(x^2+4)$

Let's do the long division

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

$color(white)(aaaa)$ $5x^4$ $color(white)(aa aaaa)$ $+20x^2$ $color(white)(aaaaaaaaaaa)$ $|$ $5x^2-3x-18$

$color(white)(aaaa)$ $0-3x^3$ $color(white)(aa a)$ $-18x^2$ $color(white)(aaaaaaa)$

$color(white)(aaaaaa)$ $-3x^3$ $color(white)(aaaa)$ $-18x^2-12x-1$ $color(white)(aaaaaaa)$

$color(white)(aaaaaaaa)$ $0$ $color(white)(aaaaa)$ $-18x^2-12x$ $color(white)(a)$ $-72$

$color(white)(aaaaaaaaaaaaaaaaa)$ $0$ $color(white)(aaa)$ $+12x$ $color(white)(aa)$ $+71$

Therefore,

$(5x^4-3x^3+2x^2-1)/(x^2+4)=5x^2-3x-18+(12x+71)/(x^2+4)$

The quotient is $=5x^2-3x-18$

The remainder is $=12x+71$

How do you divide (5x^4-3x^3+2x^2-1)div(x^2+4)(5x^4-3x^3+2x^2-1)div(x^2+4) using long division?