Solve 4(a+beta)^2-9(a-beta)^2

$$4 ( a + \beta ) ^ { 2 } - 9 ( a - \beta ) ^ { 2 }$$

Evaluate

$$\left(5a-\beta \right)\left(5\beta -a\right)$$

Solution Steps

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

$$4\left(a^{2}+2a\beta +\beta ^{2}\right)-9\left(a-\beta \right)^{2}$$

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

$$4a^{2}+8a\beta +4\beta ^{2}-9\left(a-\beta \right)^{2}$$

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

$$4a^{2}+8a\beta +4\beta ^{2}-9\left(a^{2}-2a\beta +\beta ^{2}\right)$$

Use the distributive property to multiply -9 by a^{2}-2a\beta +\beta ^{2}.

$$4a^{2}+8a\beta +4\beta ^{2}-9a^{2}+18a\beta -9\beta ^{2}$$

Combine 4a^{2} and -9a^{2} to get -5a^{2}.

$$-5a^{2}+8a\beta +4\beta ^{2}+18a\beta -9\beta ^{2}$$

Combine 8a\beta and 18a\beta to get 26a\beta .

$$-5a^{2}+26a\beta +4\beta ^{2}-9\beta ^{2}$$

Combine 4\beta ^{2} and -9\beta ^{2} to get -5\beta ^{2}.

$$-5a^{2}+26a\beta -5\beta ^{2}$$

Expand

$$-5a^{2}+26a\beta -5\beta ^{2}$$

Solution Steps

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

$$4\left(a^{2}+2a\beta +\beta ^{2}\right)-9\left(a-\beta \right)^{2}$$

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

$$4a^{2}+8a\beta +4\beta ^{2}-9\left(a-\beta \right)^{2}$$

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

$$4a^{2}+8a\beta +4\beta ^{2}-9\left(a^{2}-2a\beta +\beta ^{2}\right)$$

Use the distributive property to multiply -9 by a^{2}-2a\beta +\beta ^{2}.

$$4a^{2}+8a\beta +4\beta ^{2}-9a^{2}+18a\beta -9\beta ^{2}$$

Combine 4a^{2} and -9a^{2} to get -5a^{2}.

$$-5a^{2}+8a\beta +4\beta ^{2}+18a\beta -9\beta ^{2}$$

Combine 8a\beta and 18a\beta to get 26a\beta .

$$-5a^{2}+26a\beta +4\beta ^{2}-9\beta ^{2}$$

Combine 4\beta ^{2} and -9\beta ^{2} to get -5\beta ^{2}.

$$-5a^{2}+26a\beta -5\beta ^{2}$$