Solve -2p+0p(1-q)+0(1-p)q-1(1-p)(1-p)

$$- 2 p + 0 p ( 1 - q ) + 0 ( 1 - p ) q - 1 ( 1 - p ) ( 1 - p )$$

Evaluate

$$-p^{2}-1$$

Solution Steps

Multiply 1-p and 1-p to get \left(1-p\right)^{2}.

$$-2p+0p\left(1-q\right)+0\left(1-p\right)q-1\left(1-p\right)^{2}$$

Anything times zero gives zero.

$$-2p+0+0\left(1-p\right)q-1\left(1-p\right)^{2}$$

Anything times zero gives zero.

$$-2p+0+0-1\left(1-p\right)^{2}$$

Add 0 and 0 to get 0.

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

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

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

To find the opposite of 1-2p+p^{2}, find the opposite of each term.

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

The opposite of -2p is 2p.

$$-2p-1+2p-p^{2}$$

Combine -2p and 2p to get 0.

$$-1-p^{2}$$

Expand

$$-p^{2}-1$$

Solution Steps

Multiply 1-p and 1-p to get \left(1-p\right)^{2}.

$$-2p+0p\left(1-q\right)+0\left(1-p\right)q-1\left(1-p\right)^{2}$$

Anything times zero gives zero.

$$-2p+0+0\left(1-p\right)q-1\left(1-p\right)^{2}$$

Anything times zero gives zero.

$$-2p+0+0-1\left(1-p\right)^{2}$$

Add 0 and 0 to get 0.

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

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

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

To find the opposite of 1-2p+p^{2}, find the opposite of each term.

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

The opposite of -2p is 2p.

$$-2p-1+2p-p^{2}$$

Combine -2p and 2p to get 0.

$$-1-p^{2}$$