Solve 5(a+2)^2+12(a+b)(b-3)+4(b-3)^2

$$5 ( a + 2 ) ^ { 2 } + 12 ( a + b ) ( b - 3 ) + 4 ( b - 3 ) ^ { 2 }$$

Evaluate

$$12\left(b-3\right)\left(a+b\right)+5\left(a+2\right)^{2}+4\left(b-3\right)^{2}$$

Solution Steps

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

$$5\left(a^{2}+4a+4\right)+12\left(a+b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 5 by a^{2}+4a+4.

$$5a^{2}+20a+20+12\left(a+b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 12 by a+b.

$$5a^{2}+20a+20+\left(12a+12b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 12a+12b by b-3.

$$5a^{2}+20a+20+12ab-36a+12b^{2}-36b+4\left(b-3\right)^{2}$$

Combine 20a and -36a to get -16a.

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4\left(b-3\right)^{2}$$

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

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4\left(b^{2}-6b+9\right)$$

Use the distributive property to multiply 4 by b^{2}-6b+9.

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4b^{2}-24b+36$$

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

$$5a^{2}-16a+20+12ab+16b^{2}-36b-24b+36$$

Combine -36b and -24b to get -60b.

$$5a^{2}-16a+20+12ab+16b^{2}-60b+36$$

Add 20 and 36 to get 56.

$$5a^{2}-16a+56+12ab+16b^{2}-60b$$

Expand

$$5a^{2}+12ab-16a+16b^{2}-60b+56$$

Solution Steps

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

$$5\left(a^{2}+4a+4\right)+12\left(a+b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 5 by a^{2}+4a+4.

$$5a^{2}+20a+20+12\left(a+b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 12 by a+b.

$$5a^{2}+20a+20+\left(12a+12b\right)\left(b-3\right)+4\left(b-3\right)^{2}$$

Use the distributive property to multiply 12a+12b by b-3.

$$5a^{2}+20a+20+12ab-36a+12b^{2}-36b+4\left(b-3\right)^{2}$$

Combine 20a and -36a to get -16a.

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4\left(b-3\right)^{2}$$

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

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4\left(b^{2}-6b+9\right)$$

Use the distributive property to multiply 4 by b^{2}-6b+9.

$$5a^{2}-16a+20+12ab+12b^{2}-36b+4b^{2}-24b+36$$

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

$$5a^{2}-16a+20+12ab+16b^{2}-36b-24b+36$$

Combine -36b and -24b to get -60b.

$$5a^{2}-16a+20+12ab+16b^{2}-60b+36$$

Add 20 and 36 to get 56.

$$5a^{2}-16a+56+12ab+16b^{2}-60b$$