Solve (1/2+x)(1/2-x)+1/9-16x^2-(2/3+3x)(frac{2}{3}-3x)

$$( \frac { 1 } { 2 } + x ) ( \frac { 1 } { 2 } - x ) + \frac { 1 } { 9 } - 16 x ^ { 2 } - ( \frac { 2 } { 3 } + 3 x ) ( \frac { 2 } { 3 } - 3 x )$$

Evaluate

$$-8x^{2}-\frac{1}{12}$$

Solution Steps

Consider \left(\frac{1}{2}+x\right)\left(\frac{1}{2}-x\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 \frac{1}{2}.

$$\frac{1}{4}-x^{2}+\frac{1}{9}-16x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Add \frac{1}{4} and \frac{1}{9} to get \frac{13}{36}.

$$\frac{13}{36}-x^{2}-16x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Combine -x^{2} and -16x^{2} to get -17x^{2}.

$$\frac{13}{36}-17x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Consider \left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\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 \frac{2}{3}.

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-\left(3x\right)^{2}\right)$$

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

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-3^{2}x^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-9x^{2}\right)$$

To find the opposite of \frac{4}{9}-9x^{2}, find the opposite of each term.

$$\frac{13}{36}-17x^{2}-\frac{4}{9}+9x^{2}$$

Subtract \frac{4}{9} from \frac{13}{36} to get -\frac{1}{12}.

$$-\frac{1}{12}-17x^{2}+9x^{2}$$

Combine -17x^{2} and 9x^{2} to get -8x^{2}.

$$-\frac{1}{12}-8x^{2}$$

Expand

$$-8x^{2}-\frac{1}{12}$$

Solution Steps

Consider \left(\frac{1}{2}+x\right)\left(\frac{1}{2}-x\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 \frac{1}{2}.

$$\frac{1}{4}-x^{2}+\frac{1}{9}-16x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Add \frac{1}{4} and \frac{1}{9} to get \frac{13}{36}.

$$\frac{13}{36}-x^{2}-16x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Combine -x^{2} and -16x^{2} to get -17x^{2}.

$$\frac{13}{36}-17x^{2}-\left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\right)$$

Consider \left(\frac{2}{3}+3x\right)\left(\frac{2}{3}-3x\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 \frac{2}{3}.

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-\left(3x\right)^{2}\right)$$

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

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-3^{2}x^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{13}{36}-17x^{2}-\left(\frac{4}{9}-9x^{2}\right)$$

To find the opposite of \frac{4}{9}-9x^{2}, find the opposite of each term.

$$\frac{13}{36}-17x^{2}-\frac{4}{9}+9x^{2}$$

Subtract \frac{4}{9} from \frac{13}{36} to get -\frac{1}{12}.

$$-\frac{1}{12}-17x^{2}+9x^{2}$$

Combine -17x^{2} and 9x^{2} to get -8x^{2}.

$$-\frac{1}{12}-8x^{2}$$