How can I factor the equation $a^3 - 5a^2 + 8a -4 = 0$? Consider the equality
$$
a^3 - 5a^2 + 8a -4 = 0
$$
What are the steps to factor this into the following?
$$
(a − 1)(a − 2)^2 = 0
$$
Is there some technique to do this?
 A: You have to guess one root, then do long division and then when you are left with second order polynomial, factorize. So lets say we guessed root $a = 1$, then long division by $(a -1)$ gives $a^2-4a + 4$ which you can then factorize to $(a - 2)^2$.
Often with those types of problems(in schools), you can break out a factor of $a$ (not in this case though), and then you are left with a simple quadratic polynomial.
You could also try looking at a plot of the function and see where the roots are, and then do long division until you can do factorization of quadratic polynomial that's left.
A: \begin{align}
a^3 - 5a^2 + 8a -4 &= a^3 - a^2 -4a^2 + 8a -4 \\
                   &= a^2(a-1) - 4(a^2-2a+1) \\
                   &= a^2(a-1) - 4(a-1)^2\\
                   &= (a-1) \left[ a^2 - 4(a-1)\right] \\
                   &= (a-1)(a-2)^2
\end{align}
A: Adding to what JamMaster said, your long division guesses should not be random. Take the constant $C$ of the equation (in this case $-4$) and take the coefficient of your variable raised to the highest degree (in this case $1$). Now, take all the factors of $-4$ and all the factors of $1$ and divide the former by the latter. Your real solution(s) will be $\pm1, \pm2,$ and/or $\pm4$.
Plug in the $6$ values using synthetic division until one works. That way you don't have to arbitrarily guess roots of the equation.
There are other tests to determine how many roots are positive or negative and how many are real vs. imaginary/complex. If you want me to post these as well, I can.
A: Integer roots of this ploynomial, if any, are divisors of $-4$, i.e. are  among  $\pm 1,\pm2,\pm 4$. From the signs of the coefficients, we see there can be no negative root, hence you only have to test $1,2,4$. It happens $1$ and $2$ are roots, but $4$ is not. 
As a consequence, $p(a)=a^3-5a^2+8a-4$ is divisible by $a-1$ and $a-2$. These are coprime, s o $p(a)$ is divisible by $(a-1)(a-2)=a^2-3a+2$. Then long division gives the factorisation.
Note: As $2$ is also a root of $p'(a)=3a^2-10a+8$, this implies $2$ is a double root of $p(a)$, i.e. $p(a)$ is divisible by $(a-2)^2$, and finally by $(a-1)(a-2)^2$. If we compare leading coefficients of both, we conclude $p(a)$ is $(a-1)(a-2)^2$.
A: Adding to what  others have said, in order to find (or disprove the existence of) rational roots, there's a theorem due to Gauss (I think) why states that:
Let $f=\sum_{i=0}^na_ix^i$ be a polynomial with integer coefficients, then if$\frac{c}{d}$, with $c,d\in \mathbb{Z}$, $d\neq 0$ is a root of $f$, $c$ divides $a_0$ and $d$ divides $a_n$.
This narrows down the possibilites for trial and error.
A: L.A.F.2. is referring to a special case of the "rational root theorem":
  "Any rational root of the polynomial equation $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ must be of the form $\frac{m}{n}$ where the numerator, m, evenly divides the constant term, $a_0$, and the denominator, n, evenly divides the leading coefficient, $a_n$."   
Here, the leading coefficient is 1 so the denominator must be 1 or -1 which means any rational root must be an integer.  The constant term is 4 so any rational root must be an integer that divides 4.  That is, any rational root must be 1, -1, 2, -2, 4, or -4.  Plug each of those into the polynomial to see which, if any, satisfy the equation.  
(Of course, there is no a-priori guarantee that there are any rational roots.)
