Solve (4a+1)^2-4(a+2)(3-a)=

$$( 4 a + 1 ) ^ { 2 } - 4 ( a + 2 ) ( 3 - a ) =$$

Evaluate

$$20a^{2}+4a-23$$

Solution Steps

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

$$16a^{2}+8a+1-4\left(a+2\right)\left(3-a\right)$$

Use the distributive property to multiply -4 by a+2.

$$16a^{2}+8a+1+\left(-4a-8\right)\left(3-a\right)$$

Use the distributive property to multiply -4a-8 by 3-a and combine like terms.

$$16a^{2}+8a+1-4a+4a^{2}-24$$

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

$$16a^{2}+4a+1+4a^{2}-24$$

Combine 16a^{2} and 4a^{2} to get 20a^{2}.

$$20a^{2}+4a+1-24$$

Subtract 24 from 1 to get -23.

$$20a^{2}+4a-23$$

Expand

$$20a^{2}+4a-23$$

Solution Steps

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

$$16a^{2}+8a+1-4\left(a+2\right)\left(3-a\right)$$

Use the distributive property to multiply -4 by a+2.

$$16a^{2}+8a+1+\left(-4a-8\right)\left(3-a\right)$$

Use the distributive property to multiply -4a-8 by 3-a and combine like terms.

$$16a^{2}+8a+1-4a+4a^{2}-24$$

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

$$16a^{2}+4a+1+4a^{2}-24$$

Combine 16a^{2} and 4a^{2} to get 20a^{2}.

$$20a^{2}+4a+1-24$$

Subtract 24 from 1 to get -23.

$$20a^{2}+4a-23$$