Question

How do you divide `(-18a+3a^3+9+6a^2)div(-3+3a)` using synthetic division?

Answer 1

The remainder is `0` and the quotient is `=a^2+3a-3`

Explanation:

Divide the dividend and the divisor by `3`

Let's perform the synthetic division

`color(white)(aaaa)` `1` `|` `color(white)(aaaa)` `1` `color(white)(aaaa)` `2` `color(white)(aaaaaa)` `-6` `color(white)(aaaaa)` `3`

`color(white)(aaaaa)` `|` `color(white)(aaaa)` `color(white)(aaaaa)` `1` `color(white)(aaaaaaaa)` `3` `color(white)(aaa)` `-3`

`color(white)(aaaaaaaaa)`_________

`color(white)(aaaaaa)` `|` `color(white)(aaaa)` `1` `color(white)(aaaa)` `3` `color(white)(aaaaa)` `-3` `color(white)(aaaaa)` `color(red)(0)`

The remainder is `0` and the quotient is `=a^2+3a-3`

`(a^3+2a^2-6a+3)/(a-1)=a^2+3a-3`

Answer 2

`(3a^3+6a^2-18a+9)div(3a-3)=(a^2+3a-3 )+(0)`

Explanation:

Here ,

`(-18a+3a^3+9+6a^2)div(-3+3a)`

`=>(3a^3+6a^2-18a+9)/(3a-3)=(cancel3(a^3+2a^2-6a+3))/(cancel3(a-1)`

`=>(a^3+2a^2-6a+3)div(a-1)`

Using synthetic division :

We have , `p(a)=(a^3+2a^2-6a+3) and "divisor : " a=1`

We take , coefficients of `p(a) to 1,2,-6,3`

. `1|` `1color(white)(........)2color(white)(......)-6color(white)(........)3`
`ulcolor(white)(...)|` `ul(0color(white)( ........)1color(white)(..........)3color(white)(.....)-3`
`color(white)(......)1color(white)(........)3color(white)(....)-3color(white)(........)color(violet)(ul|0|`
We can see that , quotient polynomial :

`q(a)=a^2+3a-3 and"the Remainder"=0`

Hence ,

`(a^3+2a^2-6a+3)div(a-1)=(a^2+3a-3 )+(0)`

Multiplying numerator and denominator of LHS by `3`

`(3(a^3+2a^2-6a+3))/(3(a-1))=(a^2+3a-3 )+(0)`

`(3a^3+6a^2-18a+9)/(3a-3)=(a^2+3a-3 )+(0)`