Let x1,x2,x3,x4,x5 be the roots of the polynomial f(x)=x5+x2+1, and let g(x)=x2−2. Find g(x1)g(x2)g(x3)g(x4)g(x5).
Solution — click to reveal
Since x1,x2,x3,x4,x5 are the roots of f(x)=x5+x2+1, we can write x5+x2+1=(x−x1)(x−x2)(x−x3)(x−x4)(x−x5).Also, g(x)=x2−2=(x−2)(x+2), so g(x1)g(x2)g(x3)g(x4)g(x5)=(x1−2)(x1+2)(x2−2)(x2+2)(x3−2)(x3+2)(x4−2)(x4+2)(x5−2)(x5+2)=(x1−2)(x2−2)(x3−2)(x4−2)(x5−2)×(x1+2)(x2+2)(x3+2)(x4+2)(x5+2)=(2−x1)(2−x2)(2−x3)(2−x4)(2−x5)×(−2−x1)(−2−x2)(−2−x3)(−2−x4)(−2−x5)=f(2)f(−2)=(42+2+1)(−42+2+1)=−23.