Question

How do you divide `( 6x^3 + 10x^2 + x + 8)/(2x^2 + 1)`?

Answer 1

Long divide the coefficients to find quotient `3x+5` with remainder `-2x+3`.

Explanation:

I like to long divide the coefficients like this:

...not forgetting to include a `0` in the divisor to stand for the missing `x` term.

In this example we find the quotient is `3x+5` with remainder `-2x+3`

The process is similar to long division of numbers:

Write the dividend (`6, 10, 1, 8`) under the bar.

Write the divisor (`2, 0, 1`) to the left of the bar.

Start writing the quotient, term by term, choosing each successive term to match the leading term of the running remainder:

The first term of the quotient is `color(blue)(3)`, so that when multiplied by `2, 0, 1`, the resulting first term matches the leading `6` of the dividend.

We then write out the product `6, 0, 3` of `3` and `2, 0, 1` under the dividend and subtract it to give `10, -2`. We bring down the next term from the dividend alongside it to give our running remainder.

We choose the next term of the quotient `color(blue)(5)`, so that when multiplied by `2, 0, 1`, the resulting first term matches the leading term `10` of our running remainder.

We then write out the produce `10, 0, 5` of `5` and `2, 0, 1` under the running remainder and subtract it to give `-2, 3`.

There are no more terms to bring down from the dividend, so this is our final remainder.

Then interpret the coefficient sequences by applying them to the appropriate powers of `x` to find:

`(6x^3+10x^2+x+8)/(2x^2+1) = 3x+5+(-2x+3)/(2x^2+1)`

Or if you prefer:

`6x^3+10x^2+x+8 = (3x+5)(2x^2+1)+(-2x+3)`