How do i find the roots of this polynomial equation? The polynomial equation is: $x^4-5x^3+5x^2+5x-6=0$. 
How do i simplify this equation so that i can find its roots.
Please, can anyone teach me how to find roots of equations of degree 4 and degree 3.
 A: Hint. Look first for rational solutions: for a polynomial of degree $n$, 
$$a_n x^n+\cdots +a_1x+a_0$$
with integer coefficients, if $p/q\in\mathbb{Q}$ is a solution then $p$ divides $a_0$ and $q$ divides $a_n$.
In your case, a very "lucky" one I would say, try with the divisors of $-6$ (note that $a_n=1$) that is: $\pm 1$,$\pm2$,$\pm3$,$\pm6$.
A: .The key is  first substituting integral small values of $x$, which are factors of the constant term,  which is $6$. In this case, we try to substitute $1$. We see that if $f(x) = x^4-5x^3+5x^2+5x-6$, then: $f(1)=0$ and $f(-1)=0$. Hence, $x-1$ and $x+1$ are factors of $x^4-5x^3+5x^2+5x-6$.
Now perform polynomial division: divide $ x^4-5x^3+5x^2+5x-6$ by the product $(x-1)(x+1) = x^2-1$, and you should get $(x^2-5x+6)$. Once this is done, you can either use the quadratic formula, or the same technique as above to see that $2$ and $3$ are roots of $x^2-5x+6$, which means that $x^2-5x+6 = (x-2)(x-3)$.
Hence, $f(x) = (x+1)(x-1)(x-2)(x-3)$.
If it were a general cubic or quartic equation, then although formulas (like Cardano) are known, they are cumbersome to say the least, and very hard to work out on hand. So if you get a problem like this to solve by hand, it is most likely either an observation or hit-and-trial with the factors of the constant term (if there is no constant term, then $x$ is a factor, so divide by $x$ until you get a constant term) that is likely to give a breakthrough.
A: A quartic polynomial is always of form $$ax^4+bx^3+cx^2+de+e$$.
Now we know that in a quartic equation the product of the roots is equal to $e/a$. Here $e$ is equal to $-6$ and the coefficient of $x^4$ is also 1 so the product of the zeroes is equal to $-6$. We also know that the sum of the zeroes of a quartic polynomial is equal to $-b/a$ and since here also $a$ is $1$ the sum of the zeroes is equal to $-b$ I.e $5$. Thus we conclude that we have to find a set of integers whose sum is $5$ and product is $-6$. Now you just need to factorise $-6$ which will be $2×3×1×-1$ .Now you are very lucky that you got exactly 4 factors which will stand for the $4$ zeroes of the polynomial. Now you need to arrange these numbers in a way that you get $5$ which we get through $2+3+1-1$ thus we get the zeroes of the polynomial as $2,3,1,-1$. 
If the coefficient of $x^4$ is not 1 then this method won't be good as you might get your zeroes in fractions which will be very difficult to operate upon. But this is the fastest method where $a=1$ . You can carry out this method for all polynomials where $a=1$ and easily get the roots.
