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