Question

How do you divide `6x^3+5x^2-4x+4` by `2x+3`?

Answer 1

`(6x^3+5x^2-4x+4)/(2x+3) = 3x^2-2x+1` with remainder `1`

Explanation:

My preferred method is to write out the (attempted) factorisation term by term.

Write:

`6x^3+5x^2-4x+4 = (2x+3)(...`

The first term of the quotient must be `color(blue)(3x^2)`, so that when multiplied by `2x` gives us `6x^3`, so add that to what we have written...

`6x^3+5x^2-4x+4 = (2x+3)(color(blue)(3x^2)...`

That will sort out the term in `x^3`, but what happens for `x^2`?

Note that `3*3x^2 = 9x^2` which is `4x^2` more than what we want. So add `color(blue)(-2x)` to our quotient, which when multiplied by the leading `2x` of the divisor will give `4x^2`...

`6x^3+5x^2-4x+4 = (2x+3)(3x^2color(blue)(-2x)...`

Note that `3(-2x) = -6x`, but we want `-4x`, so we need an extra `2x`, which we can get by adding `color(blue)(1)` to the quotient:

`6x^3+5x^2-4x+4 = (2x+3)(3x^2-2xcolor(blue)(+1))...`

Finally note that `3*1 = 3`, but we want `4`, so we have to add a remainder `color(blue)(1)` to get:

`6x^3+5x^2-4x+4 = (2x+3)(3x^2-2x+1)color(blue)(+1)`

That was a little lengthy to explain, but with practice you will typically be able to write out the factorisation directly.

So we find:

`(6x^3+5x^2-4x+4)/(2x+3) = 3x^2-2x+1` with remainder `1`