Question 1026742
There must be a better trick. Let me know.
Maybe it is easier using synthetic division, but it did not sound so.
Here is my trick.
{{{x=1+c}}} where {{{c=sqrt(2)+sqrt(3)}}}
{{{P(x)=2x^4-8x^3-5x^2+26x-28}}}
{{{P(1+c)=2(1+c)x^4-8(1+c)^3-5(1+c)^2+26(1+c)-28}}}
{{{P(1+c)=2(1+4c+6c^2+4c^3+c^4)-8(1+3c+3c^2+c^3)-5(1+2c+c^2)+26(1+c)-28}}}
{{{matrix(6,11,
P(1+c),"=",2,"+",8c,"+",12c^2,"+",8c^3,"+",c^4,
" ","-",8,"-",24c,"-",24c^2,"-",8c^3," "," ",
" ","-",5,"-",10c,"-",5c^2," "," "," "," ",
" ","+",26,"+",26c," "," "," "," "," "," ",
" ","-",28," "," "," "," "," "," "," "," ",
"=","-",13,"+",0,"-",17c^2,"+",0,"+",2c^4) }}}
That is considerably simpler.
{{{c^2=(sqrt(2)+sqrt(3))^2=(sqrt(2))^2+2*sqrt(2)*sqrt(3)+(sqrt(3))^2=2+2sqrt(6)+3=5+2sqrt(6)}}}
{{{c^4=(c^2)^2=(5+2sqrt(6))^2=5^2+2*5*(2sqrt(6))+(2sqrt(6))^2=25+20sqrt(6)+2^2*(sqrt(6))^2=25+20sqrt(6)+4*6=25+20sqrt(6)+24=49+20sqrt(6)}}}
So substituting the expressiond found for {{{c^2}}} and {{{c^4}}} into
{{{P(1+sqrt(2)+sqrt(3))=P(1+c)=-13-17c^2+2c^4}}} , we get
{{{P(1+sqrt(2)+sqrt(3))=-13-17(5+2sqrt(6))+2(49+20sqrt(6))=-13-85-34sqrt(6)+98+40sqrt(6)=highlight(6sqrt(6))}}}
 
NOTE: Can someone please tell me what is the simpler way to the solution that I overlooked? I will be very, very grateful to whoever tells me.