well, using the remainder theorem, we can find the remainder by simply checking f(1), that is, for the factor of x - 1, we can rewrite it as x - 1 = 0 thus x = 1, so for f(x) when x = 1, we'd use f( 1 ) and whatever comes down the pike is the remainder of f(x) and x - 1 :).
![f(x)=4x^3-7x^2+7x-2 \\\\[-0.35em] ~\dotfill\\\\ f(1)=4(1)^3-7(1)^2+7(1)-2\implies f(1)=4-7+7-2\implies \stackrel{ remainder }{f(1)=\stackrel{ \downarrow }{2}}](https://img.qammunity.org/2024/formulas/mathematics/high-school/bgbt2wbys64rmcykr7l9bc8oos8kgl999c.png)
now, the factor theorem says that if x - 1 is really a factor of f(x), then f(1) will have a remainder of 0, that is f( 1 ) = 0, but hell, we just checked above, it wasn't 0, so is not really a factor.