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)?

2 Answers
Aug 12, 2018

(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)

Aug 12, 2018

(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)