Solve (5a+3)^2+(56-3)^2-2(a+5)

$$( 5 a + 3 ) ^ { 2 } + ( 56 - 3 ) ^ { 2 } - 2 ( a + 5 )$$

Evaluate

$$25a^{2}+28a+2808$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(5a+3\right)^{2}.

$$25a^{2}+30a+9+\left(56-3\right)^{2}-2\left(a+5\right)$$

Subtract 3 from 56 to get 53.

$$25a^{2}+30a+9+53^{2}-2\left(a+5\right)$$

Calculate 53 to the power of 2 and get 2809.

$$25a^{2}+30a+9+2809-2\left(a+5\right)$$

Add 9 and 2809 to get 2818.

$$25a^{2}+30a+2818-2\left(a+5\right)$$

Use the distributive property to multiply -2 by a+5.

$$25a^{2}+30a+2818-2a-10$$

Combine 30a and -2a to get 28a.

$$25a^{2}+28a+2818-10$$

Subtract 10 from 2818 to get 2808.

$$25a^{2}+28a+2808$$

Expand

$$25a^{2}+28a+2808$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(5a+3\right)^{2}.

$$25a^{2}+30a+9+\left(56-3\right)^{2}-2\left(a+5\right)$$

Subtract 3 from 56 to get 53.

$$25a^{2}+30a+9+53^{2}-2\left(a+5\right)$$

Calculate 53 to the power of 2 and get 2809.

$$25a^{2}+30a+9+2809-2\left(a+5\right)$$

Add 9 and 2809 to get 2818.

$$25a^{2}+30a+2818-2\left(a+5\right)$$

Use the distributive property to multiply -2 by a+5.

$$25a^{2}+30a+2818-2a-10$$

Combine 30a and -2a to get 28a.

$$25a^{2}+28a+2818-10$$

Subtract 10 from 2818 to get 2808.

$$25a^{2}+28a+2808$$