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

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

Evaluate

$$3\left(a^{2}+2\right)$$

Solution Steps

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

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

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

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

Use the distributive property to multiply 2 by a^{2}+2a+1.

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

Combine a^{2} and 2a^{2} to get 3a^{2}.

$$3a^{2}-4a+4+4a+2$$

Combine -4a and 4a to get 0.

$$3a^{2}+4+2$$

Add 4 and 2 to get 6.

$$3a^{2}+6$$

Expand

$$3a^{2}+6$$

Solution Steps

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

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

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

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

Use the distributive property to multiply 2 by a^{2}+2a+1.

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

Combine a^{2} and 2a^{2} to get 3a^{2}.

$$3a^{2}-4a+4+4a+2$$

Combine -4a and 4a to get 0.

$$3a^{2}+4+2$$

Add 4 and 2 to get 6.

$$3a^{2}+6$$