Question

3. Prove that each of the following polynomials has the stated number of (real) (a) p(x) -x* -4x3 +3x2 +2x- 1, 4 zeros;

Answer 1

p(x) = x^4 - 4x^3 + 3x^2 + 2x - 1 Here We use a result to (Descarle rule) state p(x) has 4 real zeros the number of +ve real roots of p(x) is number of sign changes in p(x) here p(x) = x^4 - 4x^3 + 3x^2 + 2x - 1 number of +ve real roots = 3 number of -ve real roots is number of sign changes in p(-x) p(-x) = x^4 + 4x^3 + 3x^2 - 2x -1 Therefore number of -ve real roots is 1 \r\n here total number of real roots is 3 + 1 = 4 p(x) is 4th degree polynomial Therefore p(x) has exactly 4 zeros by descarle's rule all 4 roots are real
by Descartes rule number of positive really roots=number of sign changes in P(x)=3,number of negative real roots of p(x)=number of sign changes in P(-x)=1.so total number of real zeros=3+1=4.hence proved