Solve 7(4+a)(a-4)+2(a-2)(a-3)+5(a+1)^2=

$$7 ( 4 + a ) ( a - 4 ) + 2 ( a - 2 ) ( a - 3 ) + 5 ( a + 1 ) ^ { 2 } =$$

Evaluate

$$14a^{2}-95$$

Solution Steps

Use the distributive property to multiply 7 by 4+a.

$$\left(28+7a\right)\left(a-4\right)+2\left(a-2\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

Use the distributive property to multiply 28+7a by a-4 and combine like terms.

$$-112+7a^{2}+2\left(a-2\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

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

$$-112+7a^{2}+\left(2a-4\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

Use the distributive property to multiply 2a-4 by a-3 and combine like terms.

$$-112+7a^{2}+2a^{2}-10a+12+5\left(a+1\right)^{2}$$

Combine 7a^{2} and 2a^{2} to get 9a^{2}.

$$-112+9a^{2}-10a+12+5\left(a+1\right)^{2}$$

Add -112 and 12 to get -100.

$$-100+9a^{2}-10a+5\left(a+1\right)^{2}$$

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

$$-100+9a^{2}-10a+5\left(a^{2}+2a+1\right)$$

Use the distributive property to multiply 5 by a^{2}+2a+1.

$$-100+9a^{2}-10a+5a^{2}+10a+5$$

Combine 9a^{2} and 5a^{2} to get 14a^{2}.

$$-100+14a^{2}-10a+10a+5$$

Combine -10a and 10a to get 0.

$$-100+14a^{2}+5$$

Add -100 and 5 to get -95.

$$-95+14a^{2}$$

Expand

$$14a^{2}-95$$

Solution Steps

Use the distributive property to multiply 7 by 4+a.

$$\left(28+7a\right)\left(a-4\right)+2\left(a-2\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

Use the distributive property to multiply 28+7a by a-4 and combine like terms.

$$-112+7a^{2}+2\left(a-2\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

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

$$-112+7a^{2}+\left(2a-4\right)\left(a-3\right)+5\left(a+1\right)^{2}$$

Use the distributive property to multiply 2a-4 by a-3 and combine like terms.

$$-112+7a^{2}+2a^{2}-10a+12+5\left(a+1\right)^{2}$$

Combine 7a^{2} and 2a^{2} to get 9a^{2}.

$$-112+9a^{2}-10a+12+5\left(a+1\right)^{2}$$

Add -112 and 12 to get -100.

$$-100+9a^{2}-10a+5\left(a+1\right)^{2}$$

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

$$-100+9a^{2}-10a+5\left(a^{2}+2a+1\right)$$

Use the distributive property to multiply 5 by a^{2}+2a+1.

$$-100+9a^{2}-10a+5a^{2}+10a+5$$

Combine 9a^{2} and 5a^{2} to get 14a^{2}.

$$-100+14a^{2}-10a+10a+5$$

Combine -10a and 10a to get 0.

$$-100+14a^{2}+5$$

Add -100 and 5 to get -95.

$$-95+14a^{2}$$