Solve $p_4(x) = x^4 −(2m + 4)x^2 + (m−2)^2 $such that $p_4$ is a product of two non-constant integer-coeficient polynomials

Question

I'm having trouble getting the starting idea for a problem I've been presented with:

I need to find values for m (integer) such that the following polynomial

$p_4(x) = x^4 −(2m + 4)x^2 + (m−2)^2$; $m$ is integer

can be described as the product of two non-constant integer-coeficient polynomials.

Any help/pointers/tips on how to go about this are greatly appreciated!

Edit: I meant polynomials with integer coeficients, but for some reason typed positive.

Answer 1

Note that $$x^2-(2m+4)x^2+(m-2)^2=(x^2-m-2)^2+(m-2)^2-(m+2)^2=(x^2-m-2)^2-8m$$

It now depends on what kind of factorisation you want. If $8m$ is a square in the relevant context then you can factorise as the difference of two squares in the usual way.

If you look carefully you will see that this is the standard technique of completing the square.

Then you say you need your factors to be positive - well each factor will be quadratic in $x$ so you can test for this in the usual way for quadratics.

The essential thing here is to notice that you can do this in two steps (often possible with quartics where the terms in $x^3$ and $x$ are missing). You first treat the variable as $x^2$ - (substitute $y=x^2$ if it helps). This reduces the problem to two expressions in $x^2$, which are handled using standard methods.

Since you are looking for positive factors, there is no point looking for linear or cubic factors, which will inevitably take both positive and negative values.

Answer 2

note that :$$ax^2+bx+c=a(x-\frac{-b+\sqrt{\Delta }}{2})(x-\frac{-b+\sqrt{\Delta }}{2})$$now apply $$y=x^2 $$so you have $$y^2-(2m+4)y+(m-2)^2\\\Delta=(2m+4)^2-4(m-2)^2=(4m^2+16+16m)-4(m^2+4-4m)=32m\\y^2-(2m+4)y+(m-2)^2=(y-\frac{+(2m+4)+\sqrt{32m}}{2})(y-\frac{+(2m+4)-\sqrt{32m}}{2})\\y=x^2\\=(x^2-\frac{+(2m+4)+\sqrt{32m}}{2})(x^2-\frac{+(2m+4)-\sqrt{32m}}{2})$$