Solve 0(1-2a)^2-(1+2a)^2+(4a+1)^2

$$0 ( 1 - 2 a ) ^ { 2 } - ( 1 + 2 a ) ^ { 2 } + ( 4 a + 1 ) ^ { 2 }$$

Evaluate

$$4a\left(3a+1\right)$$

Solution Steps

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

$$0\left(1-4a+4a^{2}\right)-\left(1+2a\right)^{2}+\left(4a+1\right)^{2}$$

Anything times zero gives zero.

$$0-\left(1+2a\right)^{2}+\left(4a+1\right)^{2}$$

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

$$0-\left(1+4a+4a^{2}\right)+\left(4a+1\right)^{2}$$

To find the opposite of 1+4a+4a^{2}, find the opposite of each term.

$$0-1-4a-4a^{2}+\left(4a+1\right)^{2}$$

Subtract 1 from 0 to get -1.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(4a+1\right)^{2}.

$$-1-4a-4a^{2}+16a^{2}+8a+1$$

Combine -4a^{2} and 16a^{2} to get 12a^{2}.

$$-1-4a+12a^{2}+8a+1$$

Combine -4a and 8a to get 4a.

$$-1+4a+12a^{2}+1$$

Add -1 and 1 to get 0.

$$4a+12a^{2}$$

Expand

$$12a^{2}+4a$$

Solution Steps

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

$$0\left(1-4a+4a^{2}\right)-\left(1+2a\right)^{2}+\left(4a+1\right)^{2}$$

Anything times zero gives zero.

$$0-\left(1+2a\right)^{2}+\left(4a+1\right)^{2}$$

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

$$0-\left(1+4a+4a^{2}\right)+\left(4a+1\right)^{2}$$

To find the opposite of 1+4a+4a^{2}, find the opposite of each term.

$$0-1-4a-4a^{2}+\left(4a+1\right)^{2}$$

Subtract 1 from 0 to get -1.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(4a+1\right)^{2}.

$$-1-4a-4a^{2}+16a^{2}+8a+1$$

Combine -4a^{2} and 16a^{2} to get 12a^{2}.

$$-1-4a+12a^{2}+8a+1$$

Combine -4a and 8a to get 4a.

$$-1+4a+12a^{2}+1$$

Add -1 and 1 to get 0.

$$4a+12a^{2}$$