Question

How do you use polynomial synthetic division to divide `(x^4-6x^2+9)div(x-sqrt3)` and write the polynomial in the form `p(x)=d(x)q(x)+r(x)`?

Answer 1

`(x^4-6x^2+9)/(x-sqrt3)=(x-sqrt3)(x^2+2sqrt3x+3)+0`

Explanation:

`(x^4-6x^2+9)/(x-sqrt3)`

`=(x^3(x-sqrt3)+sqrt3x^3-6x^2+9)/(x-sqrt3)`

`=(x^3(x-sqrt3)+sqrt3x^2(x-sqrt3)-3x^2+9)/(x-sqrt3)`

`=(x^3(x-sqrt3)+sqrt3x^2(x-sqrt3)-3x(x-sqrt3)-3sqrt3(x-sqrt3))/(x-sqrt3)`

`=x^3+sqrt3x^2-3x-3sqrt3+0`

`sqrt3` is the solution again, so `x-sqrt3` divide this polynomial.

`(x^3+sqrt3x^2-3x-3sqrt3)/(x-sqrt3)`

`=(x^2(x-sqrt3)+2sqrt3x^2-3x-3sqrt3)/(x-sqrt3)`

`=(x^2(x-sqrt3)+2sqrt3x(x-sqrt3)+3(x-sqrt3))/(x-sqrt3)`

`=x^2+2sqrt3x+3`

So:

`(x^4-6x^2+9)/(x-sqrt3)=(x-sqrt3)(x^2+2sqrt3x+3)+0`

\0/ Here's our answer !

( and we can notice that `sqrt3` is a double solution of `x^4-6x^2+9`)

Answer 2

`(x^4-6x^2+9)=(x-sqrt3)(x^3+sqrt3x^2-3x-3sqrt3 )+(0)`

Explanation:

`(x^4-6x^2+9)div(x-sqrt3)`

Using synthetic division :

We have , `p(x)=(x^4+0x^3-6x^2+0x+9) and "divisor : " x=sqrt3`

We take ,coefficients of `p(x) to 1,0,-6,0,9`

`sqrt3|` `1color(white)(.......)0color(white)(......)-6color(white)(........)0color(white)(........)9`
`ulcolor(white)(....)|` `ul(0color(white)( ....)sqrt3color(white)(........)3color(white)(...)-3sqrt3color(white)(...)-9`
`color(white)(......)1color(white)(......)sqrt3color(white)(....)-3color(white)(..)-3sqrt3color(white)(....)color(violet)(ul|0|`
We can see that , quotient polynomial :

`q(x)=x^3+sqrt3x^2-3x-3sqrt3 and"the Remainder " r(x)=0`

Hence ,

`(x^4-6x^2+9)=(x-sqrt3)(x^3+sqrt3x^2-3x-3sqrt3 )+(0)`