How do you divide $\frac{- 3 x^{4} - 9 x^{2} + 2 x - 3}{x^{2} + 4}$ $(-3x^4-9x^2+2x-3)/(x^2+4)$ ?

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How do you divide $\frac{- 3 x^{4} - 9 x^{2} + 2 x - 3}{x^{2} + 4}$ $(-3x^4-9x^2+2x-3)/(x^2+4)$ ?

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Answer 1 · 2017-04-28T16:41:51

$- 3 x^{2} + 3 + \frac{2 x - 15}{x^{2} + 4}$ $-3x^2+3+[2x-15]/[x^2+4]$

Explanation:

$\frac{000}{x^{2} + 4} \frac{000}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+4)(000)/color(black)(")"bar(-3x^4-9x^2+2x-3)))$

First, remember to fill in any powers we don't have in the divisor and dividend with a $0 x$ $0x$ to that power

$\frac{000}{x^{2} + 0 x + 4} \frac{000}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)(000)/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$

Now divide the first term in the dividend with the first term in the divisor
( $- 3 x^{4} \div x^{2})$ $-3x^4-:x^2)$
[Notice how the result from the division is written above its corresponding power $(x^{2})$ $(x^2)$ ]

$\frac{000}{x^{2} + 0 x + 4} \frac{000 - 3 x^{2}}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(000)color(black)(-3x^2)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$

Now multiply our divisor $(x^{2} + 0 x + 4)$ $(x^2+0x+4)$ with the result of our division $(- 3 x^{2})$ $(-3x^2)$ and write it like so...

$\frac{000}{x^{2} + 0 x + 4} \frac{000 - 3 x^{2}}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(000)color(black)(-3x^2)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 - 3 x^{4} - 0 x^{3} - 12 x^{2}$ $color(white)000000000000-3x^4-0x^3-12x^2$

Remember just like regular long division, we subtract the result of our multiplication so the signs change

$\frac{000}{x^{2} + 0 x + 4} \frac{000 - 3 x^{2}}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(000)color(black)(-3x^2)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$

Now we get...

$\frac{000}{x^{2} + 0 x + 4} \frac{000 - 3 x^{2}}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(000)color(black)(-3x^2)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 00000 x^{4} + 0 x^{3} + 3 x^{2}$ $color(white)00000000000bar(color(white)(0000)0x^4+0x^3+3x^2)$

Bring down our remaining terms

$\frac{000}{x^{2} + 0 x + 4} \frac{000 - 3 x^{2}}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(000)color(black)(-3x^2)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$

Now divide the first term from the subtraction with the first term from the divisor, kind of like from regular long division $(3 x^{2} \div x^{2})$ $(3x^2-:x^2)$

$\frac{000}{x^{2} + 0 x + 4} \frac{0000000 - 3 x^{2} + 3}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(0000000)color(black)(-3x^2+3)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$

and again, multiply the divisor $(x^{2} + 0 x + 4)$ $(x^2+0x+4)$ with the result from our division $(3)$ $(3)$

$\frac{000}{x^{2} + 0 x + 4} \frac{0000000 - 3 x^{2} + 3}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(0000000)color(black)(-3x^2+3)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$
$000000000000000000000000003 x^{2} + 0 x + 12$ $color(white)(00000000000000000000000000)3x^2+0x+12$

[Remember we are subtracting so don't forget to change the signs!]

$\frac{000}{x^{2} + 0 x + 4} \frac{0000000 - 3 x^{2} + 3}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(0000000)color(black)(-3x^2+3)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$
$000000000000000000000000 - 3 x^{2} - 0 x - 12$ $color(white)(000000000000000000000000)-3x^2-0x-12$

$\frac{000}{x^{2} + 0 x + 4} \frac{0000000 - 3 x^{2} + 3}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(0000000)color(black)(-3x^2+3)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$
$000000000000000000000000 - 3 x^{2} - 0 x - 12$ $color(white)(000000000000000000000000)-3x^2-0x-12$
$000000000000000000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000 x^{2} + 2 x - 15$ $color(white)(000000000000000000000000)bar(color(white)(000)0x^2+2x-15)$

$\frac{000}{x^{2} + 0 x + 4} \frac{0000000 - 3 x^{2} + 3}{) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{4} + 0 x^{3} - 9 x^{2} + 2 x - 3}$ $color(white)((000)/color(black)(x^2+0x+4)[color(white)(0000000)color(black)(-3x^2+3)]/color(black)(")"bar(-3x^4+0x^3-9x^2+2x-3)))$
$000000000000 + 3 x^{4} + 0 x^{3} + 12 x^{2}$ $color(white)000000000000+3x^4+0x^3+12x^2$
$00000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000000000003 x^{2} + 2 x - 3$ $color(white)00000000000bar(color(white)(000000000000000)3x^2)+2x-3$
$000000000000000000000000 - 3 x^{2} - 0 x - 12$ $color(white)(000000000000000000000000)-3x^2-0x-12$
$000000000000000000000000 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0000000002 x - 15$ $color(white)(000000000000000000000000)bar(color(white)(000000000)2x-15)$

Now, we are left with $2 x - 15$ $2x-15$ . Before you jump in and try to divide again, notice how we have gotten to the point where the first term in our divisor $(x^{2})$ $(x^2)$ is larger than the first term from our subtraction $(2 x)$ $(2x)$ .
This means that $- 3 x^{2} + 3$ $-3x^2+3$ is our quotient and $2 x - 15$ $2x-15$ is our remainder.

When dividing polynomials, our answer is defined as the quotient $(- 3 x^{2} + 3)$ $color(red)[(-3x^2+3)]$ . Plus the remainder $(2 x - 15)$ $color(green)[(2x-15)]$ divided by the divisor $(x^{2} + 4)$ $color(blue)[(x^2+4)]$

So...

$- 3 x^{2} + 3 + \frac{2 x - 15}{x^{2} + 4}$ $color(red)[-3x^2+3]+color(green)[2x-15]/color(blue)[x^2+4]$

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Explanation:

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