Solve f(x)=(x+1/5)^2(x-5)^3

$$f ( x ) = ( x + \frac { 1 } { 5 } ) ^ { 2 } ( x - 5 ) ^ { 3 }$$

Evaluate

$$\frac{\left(5x+1\right)^{2}\left(x-5\right)^{3}}{25}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{5}\right)^{2}.

$$\left(x^{2}+\frac{2}{5}x+\frac{1}{25}\right)\left(x-5\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-5\right)^{3}.

$$\left(x^{2}+\frac{2}{5}x+\frac{1}{25}\right)\left(x^{3}-15x^{2}+75x-125\right)$$

Use the distributive property to multiply x^{2}+\frac{2}{5}x+\frac{1}{25} by x^{3}-15x^{2}+75x-125 and combine like terms.

$$x^{5}-\frac{73}{5}x^{4}+\frac{1726}{25}x^{3}-\frac{478}{5}x^{2}-47x-5$$

Expand

$$x^{5}-\frac{73x^{4}}{5}+\frac{1726x^{3}}{25}-\frac{478x^{2}}{5}-47x-5$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{5}\right)^{2}.

$$\left(x^{2}+\frac{2}{5}x+\frac{1}{25}\right)\left(x-5\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-5\right)^{3}.

$$\left(x^{2}+\frac{2}{5}x+\frac{1}{25}\right)\left(x^{3}-15x^{2}+75x-125\right)$$

Use the distributive property to multiply x^{2}+\frac{2}{5}x+\frac{1}{25} by x^{3}-15x^{2}+75x-125 and combine like terms.

$$x^{5}-\frac{73}{5}x^{4}+\frac{1726}{25}x^{3}-\frac{478}{5}x^{2}-47x-5$$