Solve f(x)=2(x^2+4)^2(x-4)^3

$$f ( x ) = 2 ( x ^ { 2 } + 4 ) ^ { 2 } ( x - 4 ) ^ { 3 }$$

Evaluate

$$2\left(x-4\right)^{3}\left(x^{2}+4\right)^{2}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+4\right)^{2}.

$$2\left(\left(x^{2}\right)^{2}+8x^{2}+16\right)\left(x-4\right)^{3}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$2\left(x^{4}+8x^{2}+16\right)\left(x-4\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-4\right)^{3}.

$$2\left(x^{4}+8x^{2}+16\right)\left(x^{3}-12x^{2}+48x-64\right)$$

Use the distributive property to multiply 2 by x^{4}+8x^{2}+16.

$$\left(2x^{4}+16x^{2}+32\right)\left(x^{3}-12x^{2}+48x-64\right)$$

Use the distributive property to multiply 2x^{4}+16x^{2}+32 by x^{3}-12x^{2}+48x-64 and combine like terms.

$$2x^{7}-24x^{6}+112x^{5}-320x^{4}+800x^{3}-1408x^{2}+1536x-2048$$

Expand

$$2x^{7}-24x^{6}+112x^{5}-320x^{4}+800x^{3}-1408x^{2}+1536x-2048$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+4\right)^{2}.

$$2\left(\left(x^{2}\right)^{2}+8x^{2}+16\right)\left(x-4\right)^{3}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$2\left(x^{4}+8x^{2}+16\right)\left(x-4\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-4\right)^{3}.

$$2\left(x^{4}+8x^{2}+16\right)\left(x^{3}-12x^{2}+48x-64\right)$$

Use the distributive property to multiply 2 by x^{4}+8x^{2}+16.

$$\left(2x^{4}+16x^{2}+32\right)\left(x^{3}-12x^{2}+48x-64\right)$$

Use the distributive property to multiply 2x^{4}+16x^{2}+32 by x^{3}-12x^{2}+48x-64 and combine like terms.

$$2x^{7}-24x^{6}+112x^{5}-320x^{4}+800x^{3}-1408x^{2}+1536x-2048$$