I need some help to factorize this

[wide bramble]

gkeocog

[magic obsidian]

This is not factorizable

[fierce agate]

Hiw did u know that?

[magic obsidian]

I tried it

I couldnt find any value of x that would make this equation 0 and i ask ai

It also said its not factorizable

With rational numbers

[fierce agate]

May I ask u how? Like what did u do trying it?

[magic obsidian]

I see the last digit which is -3

So -3 can made from -3x1,-1x3

So the possible values of x is ussually -3 , 1, -1, 3

One of those ussually will make the equation=0

Ill show u an example of a polynominal that is factorisable working

With rational no

[fierce agate]

Wdym by usually? Isn't this work always?

[magic obsidian]

Ye

I mean that is the only ik how to factorise polynominal

[fierce agate]

Oh I see tysm, I'm going to school now cya

[magic obsidian]

An example

[fierce agate]

Tysm dude

[charred creek]

It's possible for a quartic to factor into two quadratics where neither has any roots.

[mortal tapir]

The expression is not factorable with rational numbers.

[fierce agate]

How?

[mortal tapir]

It is NOT possible for a quartic equation to factor into two quadratics where neither has any roots.

Fundamental Theorem of Algebra states that every polynomial equation of degree n has exactly n complex roots (counted with multiplicity). Since a quadratic equation can have at most two roots, two quadratics can have at most four roots. Therefore, a quartic equation must have at least one root, which means it cannot factor into two quadratics where neither has any roots.

Hope that clear things out!

[strange patrol]

Yes it is

$x^4 + 4 x^2 + 3 = (x^2+1)(x^2+3)$
And neither has any roots (in R)

[wide bramble]

Daddy_314

[charred creek]

They're taking about rational roots.

[charred creek]

You didn't need to say any of this. You could have summed up why you thought I was wrong with "all quadratics have roots in the complex plane".

[magic obsidian]

Mmm ah must have not learnt to that point yet

[charred creek]

The factor theorem says that a root of a polynomial gives you (x-root) as a factor of the polynomial. So most high school strategies to factor focus on finding a root. It's true that if a cubic doesn't have a rational root, then it doesn't factor over the rationals.

But this example above x^4 + 4 x^2 + 3 = (x^2+1)(x^2+3) shows that polynomials can still factor over the rationals even if they have no rational roots. The polynomial just has to be degree 4 or higher.

[magic obsidian]

Aaaah

I see

I understand