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

1 Answer
Aug 1, 2018

Answer:

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)#