dansingh said:
1)y^3+6y^2+11y+6=0
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.
I'll work the first one, and then see if you can use the same procedure to work the second one.
1.) Let:
$$P(y)=y^3+6y^2+11y+6$$
Now, the rational roots theorem tells us that if $P$ has any rational roots, they must be come from a list of numbers generated by taking all of the factors of the constant term and dividing them by all of the factors of the leading coefficient. Since the leading coefficient is 1, we need only look at the factors of 6. Since all of the terms in $P$ have a " sign in front of them, we know any rational roots must be negative so that the terms with odd exponents will be negative and will cancel with the positive terms with even exponents. So, we look at this list as possible candidates:
$$y\in\{-1,-2,-3,-6\}$$
Proceeding systematically, we find:
$$P(-1)=(-1)^3+6(-1)^2+11(-1)+6=-1+6-11+6=0$$ and so we know -1 is a root.
$$P(-2)=(-2)^3+6(-2)^2+11(-2)+6=-8+24-22+6=0$$ and so we know -2 is a root.
$$P(-3)=(-3)^3+6(-3)^2+11(-3)+6=-27+54-33+6=0$$ and so we know -3 is a root.
Because a cubic polynomial can have no more than 3 roots, we are done. We can now state:
$$P(y)=y^3+6y^2+11y+6=(y+1)(y+2)(y+3)$$
If we had only found 1 rational root, then we could use polynomial division to obtain the other roots from the resulting quadratic quotient (using the quadratic formula or completing the square).