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

Answer:

#(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

Answer:

#(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)#