Question

How do you divide `(2x^3 - 11x^2 + 12x + 9)/(x-3)`?

Answer 1

The remainder is `0` and the quotient is `=2x^2-5x-3`

Explanation:

Let's perform the synthetic division

`color(white)(aaaa)` `3` `|` `color(white)(aaaa)` `2` `color(white)(aaaa)` `-11` `color(white)(aaaaaa)` `12` `color(white)(aaaaa)` `9`

`color(white)(aaaaa)` `|` `color(white)(aaaa)` `color(white)(aaaaaaa)` `6` `color(white)(aaaaa)` `-15` `color(white)(aaa)` `-9`

`color(white)(aaaaaaaaa)`##_________

`color(white)(aaaaa)` `|` `color(white)(aaaa)` `2` `color(white)(aaaaa)` `-5` `color(white)(aaaaa)` `-3` `color(white)(aaaaa)` `color(red)(0)`

The remainder is `0` and the quotient is `=2x^2-5x-3`

`(2x^3-11x^2+12x+9)/(x-3)=2x^2-5x-3`

Apply the remainder theorem

When a polynomial `f(x)` is divided by `(x-c)`, we get

`f(x)=(x-c)q(x)+r`

Let `x=c`

Then,

`f(c)=0+r`

Here,

`f(x)=2x^3-11x^2+12x+9`

Therefore,

`f(3)=2*3^3-11*3^2+12*3+9`

`=54-99+36+9`

`=0`

The remainder is `=0`

Answer 2

`(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)`

Explanation:

`(2x^3-11x^2+12x+9)div(x-3)`

We can divide this polynomial by using synthetic division

We have , `p(x)=2x^3-11x^2+12x+9 and "divisor :"x=3`

We take ,coefficients of `p(x) to 2,-11,12 ,9`

. `3 |` `2color(white)(....)-11color(white)(.......)12color(white)(.......)9`
`ulcolor(white)()|` `ul(0color(white)( ..........)6color(white)(....)-15color(white)(...)-9`
`color(white)(......)2color(white)(......)-5color(white)(...)color(white)(..)-3color(white)(.....)color(violet)(ul|0|`
We can see that , quotient polynomial :

`q(x)=2x^2-5x-3 and"the Remainder"=0`

Hence ,

`(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)`