How would I factor
$\displaystyle 64x^4-768x^3 + 3392x^2 -6529x +4625 = 0$
to
$\displaystyle (4x^2-25x +37)(16x^2-92x+125) = 0$
ok, so, are you asking us to finish factoring? or how to get where its at in factoring the problem?
$\displaystyle
(4x^2-25x +37)(16x^2-92x+125) = 0
$
you can just use the quadratic formula to finish it:
$\displaystyle x= /frac {-b(+-) /sqrt {b^2 - 4ac}}{2a} $
or to see how to get to:
$\displaystyle
(4x^2-25x +37)(16x^2-92x+125) = 0
$
with
$\displaystyle
64x^4-768x^3 + 3392x^2 -6529x +4625 = 0
$
you simply go like this:
$\displaystyle /sqrt {64^4} $
then same thing with:
$\displaystyle /sqrt {3392^2} $
This might help:
Quartic equation - Wikipedia, the free encyclopedia
Hello,
I can see no other obvious way than getting two factors (ax²+bx+c)(dx²+ex+f), since if there is a complex root, there will be its conjugate as a root too. And (x-1st root)(x-conjugate) will always give an quadratic with rational coefficients.
If it's real roots, no problem.
Well, there are 3 or 4 situations.
- all roots are rational. This is not a problem, you can develop $\displaystyle 64(x-a)(x-b)(x-c)(x-d)$, where a,b,c,d are the roots.
- at least one root is complex (this is the most general case). This is not a problem either. We know that if the coefficients of a polynomial are rational (and thus integers), then if a complex number $\displaystyle z_1=x_1+i y_1$ is a root, then its conjugate $\displaystyle z_2=x_1-i y_2$ is also a root. $\displaystyle \boxed{\text{Complex roots go by } \underline{\text{pairs}}}$
So $\displaystyle (x-z_1)(x-z_2)$ divides the polynomial, that is to say $\displaystyle (x-x_1-i y_1)(x-x_1+i y_1)$ this, again, is a product of a complex number and its conjugate : $\displaystyle (x-x_1)-i y_1$ and $\displaystyle (x-x_1)+i y_1$
If you do the product, you'll get a quadratic (x is the variable) : $\displaystyle =(x-x_1)^2+y_1^2$ (rational coefficients)
It can be deduced (with a same reasoning for the 3rd and 4th roots) that any quartic with rational coefficients can be factored into a product of 2 quadratics with rational coefficients.
Since you don't know if there are complex roots or only real roots, it's better to assume there are complex roots (real numbers are complex numbers), and thus to write the original polynomial into the product of 2 quadratics.
Is it clear ?
that's cool !Thanks Moo and o i did like that techno song lol.