If $f (x) = x^{4} + 16 x^{3} + 72 x^{2} + 64 x - 129$ , find the value of $f (x - 4)$ .

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Solution

$f (x) = x^{4} + 16 x^{3} + 72 x^{2} + 64 x - 129$

$f (x - 4) = (x - 4)^{4} + 16 (x - 4)^{3} + 72 (x - 4)^{2} + 64 (x - 4) - 129$

$= (x^{4} + (\frac{4}{1}) \cdot x^{3} \cdot (- 4) + (\frac{4}{2}) \cdot x^{2} \cdot (- 4)^{2} + (\frac{4}{3}) \cdot x \cdot (- 4)^{3} + (\frac{4}{4}) \cdot x \cdot (- 4)^{4})$

$+ 16 (x^{3} + (\frac{3}{1}) \cdot x^{2} \cdot (- 4) + (\frac{3}{2}) \cdot x \cdot (- 4)^{2} + (\frac{3}{3}) \cdot (- 4)^{3})$

$+ 72 (x^{2} + (\frac{2}{1}) \cdot x \cdot (- 4) + (\frac{2}{2}) \cdot (- 4)^{2}) + 64 (x - 4) - 129$

$= (x^{4} - 4 x^{3} + 16 x^{2} - 64 x + 256) + 16 (x^{3} - 12 x^{2} + 48 x - 64) + 72 (x^{2} - 8 x + 16) + 64 (x - 4) - 129$

$= x^{4} + 12 x^{3} + 104 x^{2} + 192 x - 1$

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