Solve (1/2(f-1)^2(2)^2(-1)(2))

$$( \frac { 1 } { 2 } ( f - 1 ) ^ { 2 } ( 2 ) ^ { 2 } ( - 1 ) ( 2 ) )$$

Evaluate

$$-4\left(f-1\right)^{2}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{1}{2}\left(f-1\right)^{2}\times 2^{3}\left(-1\right)$$

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

$$\frac{1}{2}\left(f^{2}-2f+1\right)\times 2^{3}\left(-1\right)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1}{2}\left(f^{2}-2f+1\right)\times 8\left(-1\right)$$

Multiply \frac{1}{2} and 8 to get 4.

$$4\left(f^{2}-2f+1\right)\left(-1\right)$$

Multiply 4 and -1 to get -4.

$$-4\left(f^{2}-2f+1\right)$$

Use the distributive property to multiply -4 by f^{2}-2f+1.

$$-4f^{2}+8f-4$$

Expand

$$-4f^{2}+8f-4$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{1}{2}\left(f-1\right)^{2}\times 2^{3}\left(-1\right)$$

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

$$\frac{1}{2}\left(f^{2}-2f+1\right)\times 2^{3}\left(-1\right)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1}{2}\left(f^{2}-2f+1\right)\times 8\left(-1\right)$$

Multiply \frac{1}{2} and 8 to get 4.

$$4\left(f^{2}-2f+1\right)\left(-1\right)$$

Multiply 4 and -1 to get -4.

$$-4\left(f^{2}-2f+1\right)$$

Use the distributive property to multiply -4 by f^{2}-2f+1.

$$-4f^{2}+8f-4$$