Solve left(4(p-1)right)^2-4*2left(2p^2+p-3right)

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

Evaluate

$$40-40p$$

Solution Steps

Use the distributive property to multiply 4 by p-1.

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

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

$$16p^{2}-32p+16-4\times 2\left(2p^{2}+p-3\right)$$

Multiply 4 and 2 to get 8.

$$16p^{2}-32p+16-8\left(2p^{2}+p-3\right)$$

Use the distributive property to multiply -8 by 2p^{2}+p-3.

$$16p^{2}-32p+16-16p^{2}-8p+24$$

Combine 16p^{2} and -16p^{2} to get 0.

$$-32p+16-8p+24$$

Combine -32p and -8p to get -40p.

$$-40p+16+24$$

Add 16 and 24 to get 40.

$$-40p+40$$

Expand

$$40-40p$$

Solution Steps

Use the distributive property to multiply 4 by p-1.

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

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

$$16p^{2}-32p+16-4\times 2\left(2p^{2}+p-3\right)$$

Multiply 4 and 2 to get 8.

$$16p^{2}-32p+16-8\left(2p^{2}+p-3\right)$$

Use the distributive property to multiply -8 by 2p^{2}+p-3.

$$16p^{2}-32p+16-16p^{2}-8p+24$$

Combine 16p^{2} and -16p^{2} to get 0.

$$-32p+16-8p+24$$

Combine -32p and -8p to get -40p.

$$-40p+16+24$$

Add 16 and 24 to get 40.

$$-40p+40$$