f(x) = x3 + 4x2 + x − 6 real zeros
Hello,f(x)=x^3+4x²+x-6=x^3-x²+5x²-5+x-1=x²(x-1)+5(x-1)(x+1)+x-1=(x-1)(x²+5x+5+1)=(x-1)(x²+2x+3x+6)=(x-1)(x(x+2)+3(x+2))=(x-1)(x+2)(x+3)Zeros are 1,-2,-3.