Factor $4x^3-8x^2-25x+50$ completely 
Factor $4x^3-8x^2-25x+50$ completely  

The highest numbers you can take would be $1$, $2$, or $4$. Neither of those apply to all. So let's try the $x$! But the last term $50$ doesn't have an $x$ attached. Anybody want to give a small hint please.
 A: Hint: Consider the first two terms, $4x^3-8x^2$, and the last two terms, $-25x+50$, as separate polynomials, and factor each of them. See if you can then put their factorizations together somewhow.
Mouse over the grey box to reveal a spoiler:

 $$4x^3-8x^2=4x^2(x-2)\qquad -25x+50=(-25)(x-2)$$ Now use the distributive property of multiplication to give a factorization of $$(4x^3-8x^2)+(-25x+50)$$

However, even the above spoiler is not the complete answer; once you have done this, there is a further factorization that can be done.
A: Alternatively, it is clear that $x=2$ is a root so $(x-2)$ is a factor. Divide out and you are left with a quadratic to factorise.
A: Another possible approach is to use hit and trial or Rational root theorem  (if you prefer doing this more formally) to notice that $2$ is a zero of this polynomial. 
Then:

$$ 50-25 x-8 x^2+4 x^3 = 4x^2(x-2) -25(x-2) = (4x^2-25)(x-2)=(2x+5)(2x-5)(x-2) $$

A: Hint $  $ Polynomials with coefficients  $\rm\: ad, bd, cd,\ldots, ae, be, ce,\ldots\:$ have an obvious factor, e.g.
$$\rm\: (a\,x^2\! +\! b\, x\! +\! c)\,d\,x^3 + (a\,x^2\! +\! b\,x\! +\! c)\,e\ =\ (a\,x^2\! +\! b\, x\! +\! c)\,(d\,x^3\! +\! e)\,$$
A: HINT:    The natural thing to do is attack this polynomial by trial and error with integers. However before doing this you can note that any integer root has to divide $50$ since
$$ 4n^3-8n^2-25n+50 \equiv 50 \pmod n $$
This already reduces our search. Further note that when $x$ gets to large that the first term will become dominant. We can see quickly that when $x=\pm 5$ that the first term is $\pm 500$, and too large. SO we are indeed left with cases $\pm 1, \pm 2 \pm 4$. We can also see easily that the constant term will be dominant for $\pm 1$ so there is no need to check that case. Now with only 4 options trial and error is a good option.
