# Math Help - Factoring

1. ## Factoring

How would I factor

$64x^4-768x^3 + 3392x^2 -6529x +4625 = 0$

to

$(4x^2-25x +37)(16x^2-92x+125) = 0$

2. ok, so, are you asking us to finish factoring? or how to get where its at in factoring the problem?

$

(4x^2-25x +37)(16x^2-92x+125) = 0
$

you can just use the quadratic formula to finish it:

$x= /frac {-b(+-) /sqrt {b^2 - 4ac}}{2a}$

or to see how to get to:

$

(4x^2-25x +37)(16x^2-92x+125) = 0
$

with

$

64x^4-768x^3 + 3392x^2 -6529x +4625 = 0
$

you simply go like this:

$/sqrt {64^4}$

then same thing with:

$/sqrt {3392^2}$

3. Originally Posted by 11rdc11
How would I factor

$64x^4-768x^3 + 3392x^2 -6529x +4625 = 0$

to

$(4x^2-25x +37)(16x^2-92x+125) = 0$
This might help:
Quartic equation - Wikipedia, the free encyclopedia

4. 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.

5. Originally Posted by log(xy)
ok, so, are you asking us to finish factoring? or how to get where its at in factoring the problem?

$

(4x^2-25x +37)(16x^2-92x+125) = 0
$

you can just use the quadratic formula to finish it:

$x= /frac {-b(+-) /sqrt {b^2 - 4ac}}{2a}$

or to see how to get to:

$

(4x^2-25x +37)(16x^2-92x+125) = 0
$

with

$

64x^4-768x^3 + 3392x^2 -6529x +4625 = 0
$

you simply go like this:

$/sqrt {64^4}$

then same thing with:

$/sqrt {3392^2}$
Thanks i can find the solutions using the quad formula. I'm having problems figuring out how the TI89 broke the 4th degree polynomial to two 2nd degree polynomial. I'm kind of confused as to what you are doing. Can you finish working it out so i can understand, thanks.

6. Originally Posted by Moo
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.
Hi Moo, I'm still a little confused. I'm about to try the quaditic formula chop suey suggested but would still like to understand your input. Thanks Moo and o i did like that techno song lol.

7. Originally Posted by 11rdc11
Hi Moo, I'm still a little confused. I'm about to try the quaditic formula chop suey suggested but would still like to understand your input.
Well, there are 3 or 4 situations.

- all roots are rational. This is not a problem, you can develop $64(x-a)(x-b)(x-c)(x-d)$, 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 $z_1=x_1+i y_1$ is a root, then its conjugate $z_2=x_1-i y_2$ is also a root. $\boxed{\text{Complex roots go by } \underline{\text{pairs}}}$

So $(x-z_1)(x-z_2)$ divides the polynomial, that is to say $(x-x_1-i y_1)(x-x_1+i y_1)$ this, again, is a product of a complex number and its conjugate : $(x-x_1)-i y_1$ and $(x-x_1)+i y_1$
If you do the product, you'll get a quadratic (x is the variable) : $=(x-x_1)^2+y_1^2$ (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 ?

Thanks Moo and o i did like that techno song lol.
that's cool !

8. O now it makes total sense lol. Thanks Moo!!!