How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know.  Trying to factorize the numerator first then denominator, but I get no where.  Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and neither does long division. I'm pretty stuck.
The answer from wolfram is 
$(2x-1)/((2x+1)(x-2))$.
But I can't get there.
 A: By the rational root test, $4x^3+4x^2-7x+2=0$ has roots $\frac{1}{2}$ and $-2$, so that
$$
4x^3+4x^2-7x+2=(2x - 1)^2(x + 2).
$$
In the same way we see that
$$
4x^4-17x^2+4=(2x + 1)(2x - 1)(x + 2)(x - 2).
$$
Now we can form the quotient and see the result.
A: The polynomial in the denominator can be rewritten as $4t^2 - 17t + 4$ where $t = x^2$
Use the quadratic formula to find that this polynomial has roots $t_1 = 4, t_2 = \frac{1}{4}$
Since $t = x^2$, we can factor the bottom polynomial as $(x-2)(x+2)(x-\frac{1}{2})(x + \frac{1}{2})$. The roots are solutions for $x^2 = t_1$ and $x^2 = t_2$
Find out whether any of these 4 roots is also a root of the polynomial in the numerator and factorize
A: Arthur's suggestion in the comments to try the Euclidean algorithm gnawed at me, since I had never done that with polynomials.  So, in the name of research, I thought I would try it and report back:
\begin{eqnarray*}
4x^{4}-17x^{2}+4 & = & x(4x^{3}+4x^{2}-7x+2)-2(2x^{3}+5x^{2}+x-2)\\
4x^{3}+4x^{2}-7x+2 & = & 2(2x^{3}+5x^{2}+x-2)-3(2x^{2}+3x-2)\\
2x^{3}+5x^{2}+x-2 & = & x(2x^{2}+3x-2)+(2x^{2}+3x-2)\\
2x^{2}+3x-2 & = & 2x^{2}+3x-2
\end{eqnarray*}
So the greatest common factor of the two polynomials is $2x^2+3x-2$.  Doing polynomial long division, the quotients worked out to be $$\frac{2x-1}{2x^2-3x-2}$$ like everyone else got.
To be honest, it took me a lot longer than factoring with the Rational root theorem and synthetic division.  But I suppose it's a good tool to have in the kit for when the polynomials don't have pretty linear factors.
A: Have you heard about synthetic division?. You can use it to find a rational root of the numerator and then reduce it to a quadratic polynomial which is easy to factorize. 
For factorizing the denominator, let $A = x^2$, so now you have a polynomial of the form 
$4A^2 - 17A + 4$, factorize this new polynomial, one way to do it is to use the quadratic formula.
Here's a page that explains synthetic division.
https://www.purplemath.com/modules/synthdiv.htm
A: Just above the line that says GCD it shows the quotients, which you would want because of wishing to reduce the fraction
$$  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  $$ 
$$  \left(  4 x^{4}  - 17 x^{2}  + 4 \right)  $$ 
$$  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  =  \left(  4 x^{4}  - 17 x^{2}  + 4 \right)  \cdot \color{magenta}{  \left( 0 \right) } +  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  $$
$$  \left(  4 x^{4}  - 17 x^{2}  + 4 \right)  =  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  \cdot \color{magenta}{  \left(   x  - 1 \right) } +  \left(   - 6 x^{2}  - 9 x  + 6 \right)  $$
$$  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  =  \left(   - 6 x^{2}  - 9 x  + 6 \right)  \cdot \color{magenta}{  \left(   \frac{  - 2 x  + 1 }{ 3 }  \right) } +  \left( 0 \right)  $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{  \left( 0 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left( 0 \right) }{ \left( 1  \right) } $$
$$ \color{magenta}{  \left(   x  - 1 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left( 1  \right) }{ \left(   x  - 1 \right) } $$
$$ \color{magenta}{  \left(   \frac{  - 2 x  + 1 }{ 3 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  - 2 x  + 1 }{ 3 }  \right) }{ \left(   \frac{  - 2 x^{2}  + 3 x  + 2 }{ 3 }  \right) } $$
$$  \left(  2 x  - 1 \right)  \left(   \frac{  x  - 1 }{ 3 }  \right)  -  \left(  2 x^{2}  - 3 x  - 2 \right)  \left( \frac{ 1}{3 } \right)  =  \left( 1  \right)  $$
$$  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  =  \left(  2 x  - 1 \right)  \cdot \color{magenta}{  \left(  2 x^{2}  + 3 x  - 2 \right) } +  \left( 0 \right)  $$
$$  \left(  4 x^{4}  - 17 x^{2}  + 4 \right)  =  \left(  2 x^{2}  - 3 x  - 2 \right)  \cdot \color{magenta}{  \left(  2 x^{2}  + 3 x  - 2 \right) } +  \left( 0 \right)  $$
$$  \mbox{GCD} =   \color{magenta}{  \left(  2 x^{2}  + 3 x  - 2 \right) }   $$
$$  \left(  4 x^{3}  + 4 x^{2}  - 7 x  + 2 \right)  \left(   \frac{  x  - 1 }{ 3 }  \right)  -  \left(  4 x^{4}  - 17 x^{2}  + 4 \right)  \left( \frac{ 1}{3 } \right)  =  \left(  2 x^{2}  + 3 x  - 2 \right)  $$ 
