If \( x \) is so small that terms of \( x^{3} \) and higher can be ignored, show that: \[ (2+x)(1-3 x)^{5} \approx 2-29 x+165 x^{2} \]

To approximate \((2+x)(1-3x)^{5}\), we can use the binomial expansion for \((1-3x)^{5}\). According to the binomial theorem: \[ (1-3x)^{5} \approx 1 - 3 \cdot 5 x + \frac{5 \cdot 4}{2!}(-3x)^{2} \] Calculating the coefficients: \[ = 1 - 15x + \frac{20}{2} \cdot 9x^{2} = 1 - 15x + 90x^{2} \] Now substituting this back into \((2+x)(1-3x)^{5}\): \[ (2+x)(1 - 15x + 90x^{2}) \approx 2(1 - 15x + 90x^{2}) + x(1 - 15x + 90x^{2}) \] Calculating each term: \[ = 2 - 30x + 180x^{2} + x - 15x^{2} + 90x^{3} \] Ignoring the terms with \(x^{3}\) and higher, we simplify: \[ = 2 - 29x + (180x^{2} - 15x^{2}) \] \[ = 2 - 29x + 165x^{2} \] Thus, we have shown that: \[ (2+x)(1-3x)^{5} \approx 2 - 29x + 165x^{2} \]