C-02-Q133mediumsingle_mcqIf f(x)=x3−6x2+11x−6f(x) = x^3 - 6x^2 + 11x - 6f(x)=x3−6x2+11x−6, then f(x)=0f(x) = 0f(x)=0 for x=x =x=a0,1,20, 1, 20,1,2b1,2,31, 2, 31,2,3c−1,−2,−3-1, -2, -3−1,−2,−3d2,3,42, 3, 42,3,4ব্যাখ্যাFactoring, x3−6x2+11x−6=(x−1)(x−2)(x−3)x^3-6x^2+11x-6=(x-1)(x-2)(x-3)x3−6x2+11x−6=(x−1)(x−2)(x−3), so the roots are x=1,2,3x=1,2,3x=1,2,3. Each value satisfies f(x)=0f(x)=0f(x)=0.