I'd like to point out that Caylus' posting is correct using complex factors.
one may note that it could also be simplified to
[tex]\bf f(x)=(1-2x)(x+3+7i)(x+3-7i)
\\\\\\
f(x)=(1-2x)[(x+3)+(7i)][(x+3)-(7i)]
\\\\\\
f(x)=(1-2x)~~[\stackrel{\textit{difference of squares}}{(x+3)^2-(7i)^2}] \\\\\\
f(x)=(1-2x)[(x^2+6x+9)-(7^2i^2)]
\\\\\\
f(x)=(1-2x)[(x^2+6x+9)-(49\cdot -1)]
\\\\\\
f(x)=(1-2x)[(x^2+6x+9)-(-49)]
\\\\\\
f(x)=(1-2x)[(x^2+6x+9)+49]
\\\\\\
f(x)=(1-2x)(x^2+6x+58)[/tex]
