Question

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

Answer 1

`3x^2-12x+32-93/(x+3)`

Explanation:

`"one way is to use the divisor as a factor in the numerator"`

`"consider the numerator"`

`color(red)(3x^2)(x+3)color(magenta)(-9x^2)-3x^2-4x+3`

`=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(magenta)(+36x)-4x+3`

`=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(red)(+32)(x+3)color(magenta)(-96)+3`

`=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(red)(+32)(x+3)-93`

`"quotient "=color(red)(3x^2-12x+32)," remainder "=-93`

`rArr(3x^3-3x^2-4x+3)/(x+3)=3x^2-12x+32-93/(x+3)`

Answer 2

The quotient is `=3x^2-12x+32` and the remainder is `=-93`

Explanation:

We perform a long division

`color(white)(aaaa)` `x+3` `|` `color(white)(aaaa)` `3x^3-3x^2-4x+3` `color(white)(aaaa)` `|` `3x^2-12x+32`

`color(white)(aaaaaaaaaaaaaaa)` `3x^3+9x^2`

`color(white)(aaaaaaaaaaaaaaaaa)` `0-12x^2-4x`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `-12x^2-36x`

`color(white)(aaaaaaaaaaaaaaaaaaaaaa)` `+0+32x+3`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)` `+32x+96`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)` `-0-93`

Therefore,

`(3x^3-3x^2-4x+3)/(x+3)=3x^2-12x+32-93/(x+3)`