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.
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.
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
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.
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
Just above the line that says GCD it shows the quotients, which you would want because of wishing to reduce the fraction
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) $$
$$ \left( 4 x^{4} - 17 x^{2} + 4 \right) $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( 4 x^{4} - 17 x^{2} + 4 \right) \cdot \color{magenta}{ \left( 0 \right) } + \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) $$ $$ \left( 4 x^{4} - 17 x^{2} + 4 \right) = \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( - 6 x^{2} - 9 x + 6 \right) $$ $$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( - 6 x^{2} - 9 x + 6 \right) \cdot \color{magenta}{ \left( \frac{ - 2 x + 1 }{ 3 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 0 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 0 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( x - 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 2 x + 1 }{ 3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 2 x + 1 }{ 3 } \right) }{ \left( \frac{ - 2 x^{2} + 3 x + 2 }{ 3 } \right) } $$ $$ \left( 2 x - 1 \right) \left( \frac{ x - 1 }{ 3 } \right) - \left( 2 x^{2} - 3 x - 2 \right) \left( \frac{ 1}{3 } \right) = \left( 1 \right) $$ $$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( 2 x - 1 \right) \cdot \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } + \left( 0 \right) $$ $$ \left( 4 x^{4} - 17 x^{2} + 4 \right) = \left( 2 x^{2} - 3 x - 2 \right) \cdot \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } $$ $$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) \left( \frac{ x - 1 }{ 3 } \right) - \left( 4 x^{4} - 17 x^{2} + 4 \right) \left( \frac{ 1}{3 } \right) = \left( 2 x^{2} + 3 x - 2 \right) $$
