Solve (k+1){k-(3-k)}-(2-k){(1-k)-(2+k)}+1{(1-k)(3-k)-k(2+k)}

$$( k + 1 ) \{ k - ( 3 - k ) \} - ( 2 - k ) \{ ( 1 - k ) - ( 2 + k ) \} + 1 \{ ( 1 - k ) ( 3 - k ) - k ( 2 + k ) \}$$

Evaluate

$$2-4k$$

Solution Steps

To find the opposite of 3-k, find the opposite of each term.

$$\left(k+1\right)\left(k-3-\left(-k\right)\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -k is k.

$$\left(k+1\right)\left(k-3+k\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine k and k to get 2k.

$$\left(k+1\right)\left(2k-3\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of k+1 by each term of 2k-3.

$$2k^{2}-3k+2k-3-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -3k and 2k to get -k.

$$2k^{2}-k-3-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

To find the opposite of 2+k, find the opposite of each term.

$$2k^{2}-k-3-\left(2-k\right)\left(1-k-2-k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Subtract 2 from 1 to get -1.

$$2k^{2}-k-3-\left(2-k\right)\left(-1-k-k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -k and -k to get -2k.

$$2k^{2}-k-3-\left(2-k\right)\left(-1-2k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of 2-k by each term of -1-2k.

$$2k^{2}-k-3-\left(-2-4k+k+2k^{2}\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -4k and k to get -3k.

$$2k^{2}-k-3-\left(-2-3k+2k^{2}\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

To find the opposite of -2-3k+2k^{2}, find the opposite of each term.

$$2k^{2}-k-3-\left(-2\right)-\left(-3k\right)-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -2 is 2.

$$2k^{2}-k-3+2-\left(-3k\right)-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -3k is 3k.

$$2k^{2}-k-3+2+3k-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Add -3 and 2 to get -1.

$$2k^{2}-k-1+3k-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -k and 3k to get 2k.

$$2k^{2}+2k-1-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine 2k^{2} and -2k^{2} to get 0.

$$2k-1+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of 1-k by each term of 3-k.

$$2k-1+1\left(3-k-3k+k^{2}-k\left(2+k\right)\right)$$

Combine -k and -3k to get -4k.

$$2k-1+1\left(3-4k+k^{2}-k\left(2+k\right)\right)$$

Use the distributive property to multiply k by 2+k.

$$2k-1+1\left(3-4k+k^{2}-\left(2k+k^{2}\right)\right)$$

To find the opposite of 2k+k^{2}, find the opposite of each term.

$$2k-1+1\left(3-4k+k^{2}-2k-k^{2}\right)$$

Combine -4k and -2k to get -6k.

$$2k-1+1\left(3-6k+k^{2}-k^{2}\right)$$

Combine k^{2} and -k^{2} to get 0.

$$2k-1+1\left(3-6k\right)$$

Use the distributive property to multiply 1 by 3-6k.

$$2k-1+3-6k$$

Add -1 and 3 to get 2.

$$2k+2-6k$$

Combine 2k and -6k to get -4k.

$$-4k+2$$

Expand

$$2-4k$$

Solution Steps

To find the opposite of 3-k, find the opposite of each term.

$$\left(k+1\right)\left(k-3-\left(-k\right)\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -k is k.

$$\left(k+1\right)\left(k-3+k\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine k and k to get 2k.

$$\left(k+1\right)\left(2k-3\right)-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of k+1 by each term of 2k-3.

$$2k^{2}-3k+2k-3-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -3k and 2k to get -k.

$$2k^{2}-k-3-\left(2-k\right)\left(1-k-\left(2+k\right)\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

To find the opposite of 2+k, find the opposite of each term.

$$2k^{2}-k-3-\left(2-k\right)\left(1-k-2-k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Subtract 2 from 1 to get -1.

$$2k^{2}-k-3-\left(2-k\right)\left(-1-k-k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -k and -k to get -2k.

$$2k^{2}-k-3-\left(2-k\right)\left(-1-2k\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of 2-k by each term of -1-2k.

$$2k^{2}-k-3-\left(-2-4k+k+2k^{2}\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -4k and k to get -3k.

$$2k^{2}-k-3-\left(-2-3k+2k^{2}\right)+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

To find the opposite of -2-3k+2k^{2}, find the opposite of each term.

$$2k^{2}-k-3-\left(-2\right)-\left(-3k\right)-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -2 is 2.

$$2k^{2}-k-3+2-\left(-3k\right)-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

The opposite of -3k is 3k.

$$2k^{2}-k-3+2+3k-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Add -3 and 2 to get -1.

$$2k^{2}-k-1+3k-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine -k and 3k to get 2k.

$$2k^{2}+2k-1-2k^{2}+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Combine 2k^{2} and -2k^{2} to get 0.

$$2k-1+1\left(\left(1-k\right)\left(3-k\right)-k\left(2+k\right)\right)$$

Apply the distributive property by multiplying each term of 1-k by each term of 3-k.

$$2k-1+1\left(3-k-3k+k^{2}-k\left(2+k\right)\right)$$

Combine -k and -3k to get -4k.

$$2k-1+1\left(3-4k+k^{2}-k\left(2+k\right)\right)$$

Use the distributive property to multiply k by 2+k.

$$2k-1+1\left(3-4k+k^{2}-\left(2k+k^{2}\right)\right)$$

To find the opposite of 2k+k^{2}, find the opposite of each term.

$$2k-1+1\left(3-4k+k^{2}-2k-k^{2}\right)$$

Combine -4k and -2k to get -6k.

$$2k-1+1\left(3-6k+k^{2}-k^{2}\right)$$

Combine k^{2} and -k^{2} to get 0.

$$2k-1+1\left(3-6k\right)$$

Use the distributive property to multiply 1 by 3-6k.

$$2k-1+3-6k$$

Add -1 and 3 to get 2.

$$2k+2-6k$$

Combine 2k and -6k to get -4k.

$$-4k+2$$