Solve {l}{n^2-3}{+3/4a-2/3a^2}{+(3/4a-2a^2)}

$$\left. \begin{array} { l } { n ^ { 2 } - 3 } \\ { + \frac { 3 } { 4 } a - \frac { 2 } { 3 } a ^ { 2 } } \\ { + ( \frac { 3 } { 4 } a - 2 a ^ { 2 } ) } \end{array} \right.$$

Evaluate

$$-\frac{8a^{2}}{3}+n^{2}+\frac{3a}{2}-3$$

Solution Steps

Combine \frac{3}{4}a and \frac{3}{4}a to get \frac{3}{2}a.

$$n^{2}-3+\frac{3}{2}a-\frac{2}{3}a^{2}-2a^{2}$$

Combine -\frac{2}{3}a^{2} and -2a^{2} to get -\frac{8}{3}a^{2}.

$$n^{2}-3+\frac{3}{2}a-\frac{8}{3}a^{2}$$

Factor

$$\frac{6n^{2}-16a^{2}+9a-18}{6}$$

Solution Steps

Factor out \frac{1}{12}.

$$\frac{12n^{2}-36+9a-8a^{2}+9a-24a^{2}}{12}$$

Consider 12n^{2}-36+9a-8a^{2}+9a-24a^{2}. Multiply and combine like terms.

$$12n^{2}-32a^{2}+18a-36$$

Consider 12n^{2}-32a^{2}+18a-36. Factor out 2.

$$2\left(6n^{2}-16a^{2}+9a-18\right)$$

Rewrite the complete factored expression.

$$\frac{6n^{2}-16a^{2}+9a-18}{6}$$