To find the product of \((4f + 1)(f^2 + 3f + 4)\), we can use the distributive property (also known as the FOIL method for binomials). We'll distribute each term in the first expression to each term in the second expression:
\[ (4f + 1)(f^2 + 3f + 4) = 4f \cdot (f^2 + 3f + 4) + 1 \cdot (f^2 + 3f + 4) \]
Now we will distribute \(4f\):
\[ 4f \cdot f^2 = 4f^3 \] \[ 4f \cdot 3f = 12f^2 \] \[ 4f \cdot 4 = 16f \]
Next, we distribute \(1\):
\[ 1 \cdot f^2 = f^2 \] \[ 1 \cdot 3f = 3f \] \[ 1 \cdot 4 = 4 \]
Now we can combine all these terms:
\[ 4f^3 + 12f^2 + 16f + f^2 + 3f + 4 \]
Now, combine like terms:
- The \(f^3\) term: \(4f^3\)
- The \(f^2\) terms: \(12f^2 + f^2 = 13f^2\)
- The \(f\) terms: \(16f + 3f = 19f\)
- The constant term: \(4\)
Putting it all together, we get:
\[ \boxed{4f^3 + 13f^2 + 19f + 4} \]
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