Solve -25left(a+2bright)^2+9left(3a-bright)^2

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

Evaluate

$$7\left(4a-13b\right)\left(2a+b\right)$$

Solution Steps

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

$$-25\left(a^{2}+4ab+4b^{2}\right)+9\left(3a-b\right)^{2}$$

Use the distributive property to multiply -25 by a^{2}+4ab+4b^{2}.

$$-25a^{2}-100ab-100b^{2}+9\left(3a-b\right)^{2}$$

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

$$-25a^{2}-100ab-100b^{2}+9\left(9a^{2}-6ab+b^{2}\right)$$

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

$$-25a^{2}-100ab-100b^{2}+81a^{2}-54ab+9b^{2}$$

Combine -25a^{2} and 81a^{2} to get 56a^{2}.

$$56a^{2}-100ab-100b^{2}-54ab+9b^{2}$$

Combine -100ab and -54ab to get -154ab.

$$56a^{2}-154ab-100b^{2}+9b^{2}$$

Combine -100b^{2} and 9b^{2} to get -91b^{2}.

$$56a^{2}-154ab-91b^{2}$$

Expand

$$56a^{2}-154ab-91b^{2}$$

Solution Steps

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

$$-25\left(a^{2}+4ab+4b^{2}\right)+9\left(3a-b\right)^{2}$$

Use the distributive property to multiply -25 by a^{2}+4ab+4b^{2}.

$$-25a^{2}-100ab-100b^{2}+9\left(3a-b\right)^{2}$$

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

$$-25a^{2}-100ab-100b^{2}+9\left(9a^{2}-6ab+b^{2}\right)$$

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

$$-25a^{2}-100ab-100b^{2}+81a^{2}-54ab+9b^{2}$$

Combine -25a^{2} and 81a^{2} to get 56a^{2}.

$$56a^{2}-100ab-100b^{2}-54ab+9b^{2}$$

Combine -100ab and -54ab to get -154ab.

$$56a^{2}-154ab-100b^{2}+9b^{2}$$

Combine -100b^{2} and 9b^{2} to get -91b^{2}.

$$56a^{2}-154ab-91b^{2}$$