Solving $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ Could someone please explain how to solve this:
$$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$
not the answer only, but a step-by-step solution.
I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.
Thank you so much in advance!
 A: $$x^4-10x^3 +25x^2-(4x^2-40x+100)=0$$
$$\Rightarrow x^2(x^2-10x+25)-4(x^2-10x+25)=0 $$
$$\Rightarrow (x^2-4)(x^2-10x+25)=0 $$
$$\Rightarrow (x-2)(x+2)(x-5)^2=0 $$
$$\Rightarrow x_1=-2 , x_2=2 , x_{3,4}=5$$
A: First note that $x= \pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows:
$$ x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$ 
$$\qquad \Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$
$$ (x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0 $$
$$\qquad \Rightarrow (x^2-4)(x^2-10x+25)=0 $$
Hence we can solve it as 
$$ (x^2-4)(x-5)^2 =0 $$ or $$x=\{ -2,+2,5 \}$$
A: The idea is to represent the polynomial $p(x) = x^4 - 10x^3 + 21x^2 + 40x - 100$ as a product of simple factors. By Vietta formula the free terms equals to the product of roots. Thus, if we confine ourselves to integer roots, they should divide $100$. Start with $x=2$, and verify that $p(2) = 16 - 10 \times 8 + 21 \times 4 + 40 \times 2 - 100 = 16 -80 + 84 - 80 + 100 = 0$. 
Now applying long division we compute 
$$
   p(x) = (x-2)\left( x^3 - 8 x^2 + 5 x + 50 \right)
$$
Now try $x=5$. $5^2(5-8) + 5 ( 5+ 10) = 5( -15 + 15) = 0$. Apply long division again:
$$
   p(x) = (x-2)(x-5) \left(x^2 - 3 x - 10\right) = (x-2)(x+2)(x-5)^2 
$$
A: Since the polynomial has integer coefficients, the rational root theorem applies. Thus any rational root must be of the form $x=\pm p/q$, where $p$ divides the constant term 100 and $q$ divides the leading coefficient 1. In this case, the only possibility for $q$ is 1. This tells you that any rational root must be a divisor of $100=2^2*5^2$. It turns out that this polynomial does have rational roots, after which you find one you can perform polynomial division to get a complete factorization.
For instance, we have the potential rational roots $x=\pm2,\pm5,\pm10,\pm20\pm25,\pm50,\pm100$. We could plug in $x=5$ and verify that this is a root. Then,
$$
\frac{x^4 - 10x^3 + 21x^2 + 40x - 100}{x-5} = x^3-5x^2-4x+20.
$$
Since all the roots are rational, repeating this process will generate all of them. Not every polynomial with integer coefficients has rational roots (for instance $x^2-2=0$), so this won't always be the case.
