How do you divide $(4x^4 -12 x^2-x-20)/((x^2 + 4) )$ ?

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Answer 1 · 2018-07-03T05:39:49

The quotient is $(4x^2-28)$ and the remainder is $(-x+92)$

Perform a long division.

$color(white)(aaaa)$ $4x^4-12x^2-x-20$ $color(white)(aaaa)$ $|$ $x^2+4$

$color(white)(aaaa)$ $4x^4+16x^2$ $color(white)(aaaaaaaaaaaa)$ $|$ $4x^2-28$

$color(white)(aaaaaa)$ $0-28x^2-x-20$

$color(white)(aaaaaaaa)$ $-28x^2-0x-112$

$color(white)(aaaaaaaaaa)$ $+0-x+92$

The quotient is $(4x^2-28)$ and the remainder is $(-x+92)$

$(4x^4-12x^2-x-20)/(x^2+4)=4x^2-28+(-x+92)/(x^2+4)$

How do you divide (4x^4 -12 x^2-x-20)/((x^2 + 4) )(4x^4 -12 x^2-x-20)/((x^2 + 4) )?