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 $E$;
\(\ds x\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \sum_{j \mathop = 0}^n a_j \times 1^j\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \sum_{j \mathop = 0}^n a_j\) | \(=\) | \(\ds 0\) |
$\blacksquare$