Any time you divide by a number (being a potential root of the polynomial) and get a zero remainder in the synthetic division, this means that the number is indeed a root, and thus "x minus the number" is a factor.
Using the remainder's theorem, let's figure out if x-3 is a factor
[tex]\begin{gathered} x-3=0 \\ x=0+3 \\ x=3 \end{gathered}[/tex]
The polynomial is given below as
[tex]p(x)=-2x^4+4x^3+5x^2+9[/tex]
substitute the value of x=3 in the polynomial above to get
[tex]\begin{gathered} P(x)=-2x^4+4x^3+5x^2+9 \\ P(3)=-2(3)^4+4(3)^3+5(3)^2+9 \\ P(3)=-2(81)+4(27)+5(9)+9 \\ P(3)=-162+108+45+9 \\ P(3)=-162+162 \\ P(3)=0 \end{gathered}[/tex]
Hence,
P( 3) = 0
x-3 is a factor of P(x)
