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1. Oct 27, 2015

### aikismos

Just to double check, but if one wanted to, like in partial fraction decomposition, associate literal coefficients of polynomials with corresponding unknowns on the other side of the equation, the justification for this action is the definition of equality of polynomials?

EDIT: I know this isn't true. Let's see, $a - b = 2$ is as far as it can be reduced.
$3x^2 + bx = ax^2 + 5x \rightarrow a = 3, b = 5$

Another related question is how do I express symbolically that in:

$P(x) = ax^2 + bx$

P(x) has no constant term?

I'm kinda groping around for a rigorous way to express and justify the pairing of coefficients if you were to write them as two equivalent n-tuples.

Last edited: Oct 27, 2015
2. Oct 27, 2015

### Geofleur

I would put everything on the same side and group similar powers of $x$ like this: $(3-a)x^2+(b-5)x = 0$. Since this is true for all $x$, and since $x$ and $x^2$ are independent functions (their Wronskian does not vanish), their coefficients must separately each be zero. This is a lot like having, for arrow vectors in the plane, that $(3-a)\mathbf{i}+(b-5)\mathbf{j} = 0$. The only way to add a vector in the $\mathbf{i}$ direction to one in the $\mathbf{j}$ direction and get zero is if, in fact, both vectors are the zero vector.

For the second question, how about writing it as $P(0) = 0$?

3. Oct 27, 2015

### Staff: Mentor

In partial fractions decomposition, you're setting up an equation that is true for all values of the independent variable (except those that make any denominator zero). IOW, two expressions that are identically equal.
So, for $3x^2 + bx = ax^2 + 5x$, for all x, it must be the case that a = 3 and b = 5.

4. Oct 27, 2015

Thanks guys!