Solve (1/2a+3b)(1/2a-3b)+3b^2

$$( \frac { 1 } { 2 } a + 3 b ) ( \frac { 1 } { 2 } a - 3 b ) + 3 b ^ { 2 }$$

Evaluate

$$\frac{a^{2}}{4}-6b^{2}$$

Solution Steps

Consider \left(\frac{1}{2}a+3b\right)\left(\frac{1}{2}a-3b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{2}a\right)^{2}-\left(3b\right)^{2}+3b^{2}$$

Expand \left(\frac{1}{2}a\right)^{2}.

$$\left(\frac{1}{2}\right)^{2}a^{2}-\left(3b\right)^{2}+3b^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

Expand \left(3b\right)^{2}.

$$\frac{1}{4}a^{2}-3^{2}b^{2}+3b^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{4}a^{2}-9b^{2}+3b^{2}$$

Combine -9b^{2} and 3b^{2} to get -6b^{2}.

$$\frac{1}{4}a^{2}-6b^{2}$$

Expand

$$\frac{a^{2}}{4}-6b^{2}$$

Solution Steps

Consider \left(\frac{1}{2}a+3b\right)\left(\frac{1}{2}a-3b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{1}{2}a\right)^{2}-\left(3b\right)^{2}+3b^{2}$$

Expand \left(\frac{1}{2}a\right)^{2}.

$$\left(\frac{1}{2}\right)^{2}a^{2}-\left(3b\right)^{2}+3b^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

Expand \left(3b\right)^{2}.

$$\frac{1}{4}a^{2}-3^{2}b^{2}+3b^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1}{4}a^{2}-9b^{2}+3b^{2}$$

Combine -9b^{2} and 3b^{2} to get -6b^{2}.

$$\frac{1}{4}a^{2}-6b^{2}$$