For 1st am getting 6 roots
Don't have the answer bro...share solution...
@Raunak_2020 try this..
For first try to make graph of function let
f(x) has x_1,x_2,x_3 root
f(x)=x_1 nd two more root imagining then after
x_1 will be constant u know graph of f(x)
No of time it cuts u will get the solutions
Its correct @akash_2020
For second also try to make graph
@akash_2020, @Sneha_2021 what will we do if the cubic is not getting factorised....like here we knew about one of the factors being 2 so we could easily locate the other 2 roots. ..
Say for example x^3 - 3x +1
Ivt theoram will give u idea of roots (jee domain theoram)
Other another way is descartes theoram of polynomial (rmo type) no need for jee but telling one aspect of telling roots
@Sneha_2021 for x^3-3x+1=0, number of different real solutions of f (f (x)) =0 should be 7?? Pls check..
One of my roots is between 0 and 1 and the other 2 between (1,infinity) and (-infinity,-1)..
Yes 7 is correct