Solve -25(a+2b)^2+9(2a-b)^2

$$- 25 ( a + 2 b ) ^ { 2 } + 9 ( 2 a - b ) ^ { 2 }$$

Evaluate

$$\left(a-13b\right)\left(11a+7b\right)$$

Solution Steps

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

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

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

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

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

$$-25a^{2}-100ab-100b^{2}+9\left(4a^{2}-4ab+b^{2}\right)$$

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

$$-25a^{2}-100ab-100b^{2}+36a^{2}-36ab+9b^{2}$$

Combine -25a^{2} and 36a^{2} to get 11a^{2}.

$$11a^{2}-100ab-100b^{2}-36ab+9b^{2}$$

Combine -100ab and -36ab to get -136ab.

$$11a^{2}-136ab-100b^{2}+9b^{2}$$

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

$$11a^{2}-136ab-91b^{2}$$

Expand

$$11a^{2}-136ab-91b^{2}$$

Solution Steps

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

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

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

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

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

$$-25a^{2}-100ab-100b^{2}+9\left(4a^{2}-4ab+b^{2}\right)$$

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

$$-25a^{2}-100ab-100b^{2}+36a^{2}-36ab+9b^{2}$$

Combine -25a^{2} and 36a^{2} to get 11a^{2}.

$$11a^{2}-100ab-100b^{2}-36ab+9b^{2}$$

Combine -100ab and -36ab to get -136ab.

$$11a^{2}-136ab-100b^{2}+9b^{2}$$

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

$$11a^{2}-136ab-91b^{2}$$