Let r,s, and t be the roots of the equation 4x3−59x2+32x−32=0. Find the value of f(r)+f(s)+f(t), where f(x)=4x3−59x2.
Solution — click to reveal
Since r is a root of 4x3−59x2+32x−32=0, we have 4r3−59r2+32r−32=0. Thus, f(r)=4r3−59r2=32−32r.Similarly, f(s)=32−32s and f(r)=32−32r. Then f(r)+f(s)+f(t)=96−32(r+s+t)=96−32(459)=−376.