Finding coefficients for reducibility (Abstract Algebra)

[RJLiberator]

Homework Statement

Find all real numbers k such that x^2+kx+k is reducible in ℝ[x].

Homework Equations

The Attempt at a Solution

This seems like it is simple, but it is new to me so I am looking for confirmation.

We know we can find the roots of a polynomial with b^2-4ab. We want b^2-4ab to be greater than 0 for it to have roots.

But we note:

Here, a = 1, b = b, c = b.

When we have b^2-4*b = 0 we get the answers either b = 0 or b = 4.
Both of these are solutions to the problem question.

If we let b(b-4) > 0 then we get a whole array of numbers that are not solutions to this problem.

So I am stating that b =0 and b = 4 are the only solutions to the problem as they are the only solutions to b^2-4ab = 0. But why is this so?

 

[fresh_42]

Homework Statement

Find all real numbers k such that x^2+kx+k is reducible in ℝ[x].

Homework Equations

The Attempt at a Solution

This seems like it is simple, but it is new to me so I am looking for confirmation.

We know we can find the roots of a polynomial with b^2-4ab. We want b^2-4ab to be greater than 0 for it to have roots.

Greater or equal 0. Only negative solutions will lead to negative square roots.

But we note:

Here, a = 1, b = b, c = b.

When we have b^2-4*b = 0 we get the answers either b = 0 or b = 4.
Both of these are solutions to the problem question.

Right.

If we let b(b-4) > 0 then we get a whole array of numbers that are not solutions to this problem.

Why? What happens if b=6? (I assume b=k.)

So I am stating that b =0 and b = 4 are the only solutions to the problem as they are the only solutions to b^2-4ab = 0. But why is this so?

See previous comment. Plus: what happens if b=2?

 

[RJLiberator]

Fresh_42, If we have b = 6
then we have:
x^2+6x+6

This is not reducible under the real numbers, am I right?

 

[fresh_42]

No. You can see it if you draw the function, it crosses the x-axis twice which means it is 0 at these points. What would be the solutions? What has $b^2-4ab$ to do with the roots?

 

[RJLiberator]

https://www.wolframalpha.com/input/?i=x^2+6x+6https://www5b.wolframalpha.com/Calculate/MSP/MSP3831i8h4fi8e87f9e4e00000i8261g2b6d6gh03?MSPStoreType=image/gif&s=47

Is the solution then.

Okay, so you convinced me that since b^2-4ac >= 0 then for all b that satisfy this we have reducibility. (I should have known this, it's been a longe week)

That seems to answer the original thread question.

With that being stated, there is something more here:
What has b^2-4ac have to do with the roots?

That is the interesting question.
Definition of reducible is if we let f(x) exist in R(x) then f(x) is reducible if there exists g(x) and h(x) in R(x) such that f(x) = g(x)*h(x) and deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x))

And so, there clearly exists a g(x) and h(x) by the b^2-4ac relation.

 

[fresh_42]

The two roots of $ax^2+bx+c=0$ are $\frac{-b±\sqrt{b^2-4ac}}{2a}$.
So $b^2-4ac=0$ determines exactly one (double counted) root, $b^2-4ac<0$ gives complex roots and $b^2-4ac>0$ gives two real roots. Here we have $b^2-4ac=k^2-4k=k(k-4)≥0$ which is true if both factors are positive or both factors are negative.

 

[RJLiberator]

I agree, mate.

And by the definition of reducibility, we see that the roots existing means that the function is reducible for those roots.

Thank you kindly for your help on this problem.