Solve 4(a+2p)^2-9(2a-p)^2quad

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

Evaluate

$$\left(7p-4a\right)\left(p+8a\right)$$

Solution Steps

Use binomial theorem \left(m+n\right)^{2}=m^{2}+2mn+n^{2} to expand \left(a+2p\right)^{2}.

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

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

$$4a^{2}+16ap+16p^{2}-9\left(2a-p\right)^{2}$$

Use binomial theorem \left(m-n\right)^{2}=m^{2}-2mn+n^{2} to expand \left(2a-p\right)^{2}.

$$4a^{2}+16ap+16p^{2}-9\left(4a^{2}-4ap+p^{2}\right)$$

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

$$4a^{2}+16ap+16p^{2}-36a^{2}+36ap-9p^{2}$$

Combine 4a^{2} and -36a^{2} to get -32a^{2}.

$$-32a^{2}+16ap+16p^{2}+36ap-9p^{2}$$

Combine 16ap and 36ap to get 52ap.

$$-32a^{2}+52ap+16p^{2}-9p^{2}$$

Combine 16p^{2} and -9p^{2} to get 7p^{2}.

$$-32a^{2}+52ap+7p^{2}$$

Expand

$$7p^{2}+52ap-32a^{2}$$

Solution Steps

Use binomial theorem \left(m+n\right)^{2}=m^{2}+2mn+n^{2} to expand \left(a+2p\right)^{2}.

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

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

$$4a^{2}+16ap+16p^{2}-9\left(2a-p\right)^{2}$$

Use binomial theorem \left(m-n\right)^{2}=m^{2}-2mn+n^{2} to expand \left(2a-p\right)^{2}.

$$4a^{2}+16ap+16p^{2}-9\left(4a^{2}-4ap+p^{2}\right)$$

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

$$4a^{2}+16ap+16p^{2}-36a^{2}+36ap-9p^{2}$$

Combine 4a^{2} and -36a^{2} to get -32a^{2}.

$$-32a^{2}+16ap+16p^{2}+36ap-9p^{2}$$

Combine 16ap and 36ap to get 52ap.

$$-32a^{2}+52ap+16p^{2}-9p^{2}$$

Combine 16p^{2} and -9p^{2} to get 7p^{2}.

$$-32a^{2}+52ap+7p^{2}$$