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Use factor and remainder theorems to test polynomial roots.

For a polynomial p(x), (x−a) is a factor exactly when p(a)=0; the remainder on division by (x−a) is p(a). The coefficients and roots are linked by Vieta relationships.

Corrected worked example

For $p(x)=x^3-4x^2+x+6$ , $p(2)=0$ , so $x-2$ is a factor. Division gives $p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$ . The roots are $2$ , $3$ and $-1$ ; none is repeated.

For this cubic, the roots sum to $2+3-1=4=-(-4)/1$ and their product is $2\cdot3\cdot(-1)=-6=(-1)^3(6/1)$ . Use the factor test at the candidate value, then verify the full factorization and Vieta relations.

A zero of p is a number a with p(a)=0; x=0 is not the same statement as the factor x.

For $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0=0$ , the sum of all roots (with multiplicity) is $-a_{n-1}/a_n$ and their product is $(-1)^na_0/a_n$ . A repeated root requires a repeated factor; one successful factor test alone does not prove multiplicity.