Solve (2a-5)(-5-2a)-3(a-1)^2

$$( 2 a - 5 ) ( - 5 - 2 a ) - 3 ( a - 1 ) ^ { 2 }$$

Evaluate

$$22+6a-7a^{2}$$

Solution Steps

Consider \left(2a-5\right)\left(-5-2a\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square -5.

$$25-\left(2a\right)^{2}-3\left(a-1\right)^{2}$$

Expand \left(2a\right)^{2}.

$$25-2^{2}a^{2}-3\left(a-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$25-4a^{2}-3\left(a-1\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-1\right)^{2}.

$$25-4a^{2}-3\left(a^{2}-2a+1\right)$$

Use the distributive property to multiply -3 by a^{2}-2a+1.

$$25-4a^{2}-3a^{2}+6a-3$$

Combine -4a^{2} and -3a^{2} to get -7a^{2}.

$$25-7a^{2}+6a-3$$

Subtract 3 from 25 to get 22.

$$22-7a^{2}+6a$$

Expand

$$22+6a-7a^{2}$$

Solution Steps

Consider \left(2a-5\right)\left(-5-2a\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square -5.

$$25-\left(2a\right)^{2}-3\left(a-1\right)^{2}$$

Expand \left(2a\right)^{2}.

$$25-2^{2}a^{2}-3\left(a-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$25-4a^{2}-3\left(a-1\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-1\right)^{2}.

$$25-4a^{2}-3\left(a^{2}-2a+1\right)$$

Use the distributive property to multiply -3 by a^{2}-2a+1.

$$25-4a^{2}-3a^{2}+6a-3$$

Combine -4a^{2} and -3a^{2} to get -7a^{2}.

$$25-7a^{2}+6a-3$$

Subtract 3 from 25 to get 22.

$$22-7a^{2}+6a$$