Question

How do you use synthetic division to show that `x=1+sqrt3` is a zero of `x^3-3x^2+2=0`?

Answer 1

Remainder `=0=>x=1+sqrt3` is a zero of `(x^3-3x^2+2)`

Explanation:

Using synthetic division :

`diamond(x^3-3x^2+2)div(x-1-sqrt3)`

We have , `p(x)=x^3-3x^2+0x+2 and "divisor :"x=1+sqrt3`

We take ,coefficients of `p(x) to 1,-3,0,2`

`1+sqrt3 |` `1color(white)(...........)-3color(white)(...........)0color(white)(...............)2`
`ulcolor(white)(..........)|` `ul(0color(white)( .......)1+sqrt3color(white)(.......)1-sqrt3color(white)(......)-2`
`color(white)(.............)1color(white)(...)-2+sqrt3color(white)(.....)1-sqrt3color(white)(........)color(violet)(ul|0|`
We can see that , quotient polynomial :

`q(x)=x^2+(-2+sqrt3)x+1-sqrt3 and"the Remainder"=0`

Hence ,

`(x^3-3x^2+2)`=`(x-1-sqrt3)[x^2+(-2+sqrt3)x+1-sqrt3]`

Remainder `=0=>x=1+sqrt3` is a zero of `(x^3-3x^2+2)`