Yes, yes Mentock. I will include the constant because you will it to be: )
Fresh stumper…
Can anyone factor x^4 + 2x^2 + 1. This is a wee bit simpler than the one we just worked on.
Over what field?
Since this is not an equation (no equals sign) that’s not an issue. I suppose it’s obvious that this cannot have an real zeros.
The field still matters. The polynomial has 0 roots in R, 4 in C, and 12 in H, and 28 in O. So which is it?
Let’s keep it simple. Keep it in real numbers.
It may not have any real roots but it can be factored, into (x^2+1)^2. My 7th graders warm up with that one.
But what’s O, the octonions?
You can’t do this.
For X = -7, this would result in division by zero in the original statement [ 56/( X+7 ) ].
For X = -1, P( X+1 ) = 0, which is also a divide by zero error [ P( 2X )/P( X+1 ) ].
Even though the rewritten form doesn’t have a dvision by zero issue, the definitional statement is undefined, and therfore you can’t use it.
Sorry 
Yeah, it’s an intermediate algebra question for sure. A lot of people don’t see the substitution of x^4 = y^2 which then it turns into a standard y^2 + 2y + 1 expression that’s instantly factorable.
We didn’t substitute into the original statement, only in the rewritten form. Since we know P is a polynomial, the transformation from the original to the rewritten form is allowed, and the rewritten form is a true statement for all x.