find the sum of squares of roots of an equation x^3-4x^2+x+c=0

Question

using P-form

Answer 1

Let the roots be w, y, z.

(x-w)(x-y)(x-z)=(x-w)(x²-x(y+z)+yz)=x³-x²(y+z)+xyz-wx²+wx(y+z)-wyz=

x³-x²(w+y+z)+x(yz+wy+wz)-wyz=x³-4x²+x+c=0.

Therefore, w+y+z=4, yz+wy+wz=1, c=-wyz, by matching coefficients.

4²=16=(w+y+z)²=(w+(y+z))²=w²+2w(y+z)+(y+z)²=w²+2w(y+z)+y²+2yz+z²=

w²+y²+z²+2(wy+wz+yz)=w²+y²+z²+2.

Therefore, w²+y²+z²+2=16, w²+y²+z²=14, which is the sum of the squares of the roots.