Not Just Vieta's Derivatives!
P ( x ) Q ( x ) = = x 6 − x 5 − x 3 − x 2 − x x 4 − x 3 − x 2 − 1
Consider the polynomials above.
Given that z 1 , z 2 , z 3 , and z 4 are the roots of Q ( x ) = 0 , find P ( z 1 ) + P ( z 2 ) + P ( z 3 ) + P ( z 4 ) .
The answer is 6.
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2 solutions
Same method+1
Same method but I used Newton's sums.
U stole my method U thief. I will not leave U. Joking xD. Great Job !
P(x)=x^2(x^4-x^3-x-1)-x
P(x)=x^2(x^2-x)-x
P(x)=x^4-x^3-x
P(x) = x^2-x+1
In various places I have substituted x^4-x^3=x^2+1 as Q(x) is zero.
Now sum of squares of roots of Q(x)=3
Sum if roots =1
Hence answer = 3-1+4=6.
This is the simplest approach i have .
So isn't the answer 3? 3 -1 + 1? Where did you get that 4 there in the last equation?