Question

How do you long divide `(4x^4 + 4x^3 -8x + 2)/( 2x^2 - 3x + 1)`?

Answer 1

Long divide coefficients to find:

`(4x^4+4x^3-8x+2)/(2x^2-3x+1)`

`=(2x^2+5x+13/2) + (13/2x-9/2)/(2x^2-3x+1)`

Explanation:

I like to just long divide the coefficients, not forgetting to include `0`'s for missing powers of `x`. In addition, I'm not keen on fractions, so I will choose to multiply the numerator by `2` first, then divide by `2` at the end.

`2 xx (4x^4+4x^3-8x+2) = 8x^4+8x^3+0x^2-16x+4`

This process is similar to long division of decimal numbers.

Write the dividend `(8, 8, 0, -16, 4)` under the bar and the divisor `(2, -3, 1)` to the left of the bar.

Choose the first term `color(blue)(4)` of the quotient so that when multiplied by the divisor it matches the leading term of the dividend.

Write the product `(8, -12, 4)` under the dividend and subtract it to get a remainder. Bring down the next term `-16` of the dividend alongside it to give the running remainder.

Choose the next term `color(blue)(10)` of the quotient so that when multiplied by the divisor it matches the first term of the running remainder.

Write the product `(20, -30, 10)` under the running remainder and subtract it to get the new running remainder. Bring down the next term `4` of the dividend alongside it.

Choose the final term `color(blue)(13)` of the quotient as before, write down the product `(26, -39, 13)` and subtract it to give the final remainder `color(red)("("13, -9")")`.

So:

`8x^4+8x^3-16x+4 = (2x^2-3x+1)(4x^2+10x+13)+(13x-9)`

Dividing by `2`:

`4x^4+4x^3-8x+2`

`= (2x^2-3x+1)(2x^2+5x+13/2) + (13/2x-9/2)`

In other words:

`(4x^4+4x^3-8x+2)/(2x^2-3x+1)`

`=(2x^2+5x+13/2) + (13/2x-9/2)/(2x^2-3x+1)`