Coefficients of Polynomial add to 0 iff 1 is a Root

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Theorem

Let $\map E x$ be the equation in $x$ represented as:

$\ds \sum_{j \mathop = 0}^n a_j x^j = 0$

where the $a_j$s are constants.


Then $x$ is a root of $\map E x$ if and only if:

$\ds \sum_{j \mathop = 0}^n a_j = 0$

That is, $x$ is a root of $\map E x$ if and only if all the coefficients of the polynomial in $x$ sum to zero.


Proof

Letting $x = 1$ in $E$;

\(\ds x\) \(=\) \(\ds 1\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sum_{j \mathop = 0}^n a_j \times 1^j\) \(=\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sum_{j \mathop = 0}^n a_j\) \(=\) \(\ds 0\)

$\blacksquare$