Solve (3-m)^2+(-2/3m+2)^2

$$( 3 - m ) ^ { 2 } + ( - \frac { 2 } { 3 } m + 2 ) ^ { 2 }$$

Evaluate

$$\frac{13\left(3-m\right)^{2}}{9}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-m\right)^{2}.

$$9-6m+m^{2}+\left(-\frac{2}{3}m+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{2}{3}m+2\right)^{2}.

$$9-6m+m^{2}+\frac{4}{9}m^{2}-\frac{8}{3}m+4$$

Combine m^{2} and \frac{4}{9}m^{2} to get \frac{13}{9}m^{2}.

$$9-6m+\frac{13}{9}m^{2}-\frac{8}{3}m+4$$

Combine -6m and -\frac{8}{3}m to get -\frac{26}{3}m.

$$9-\frac{26}{3}m+\frac{13}{9}m^{2}+4$$

Add 9 and 4 to get 13.

$$13-\frac{26}{3}m+\frac{13}{9}m^{2}$$

Expand

$$\frac{13m^{2}}{9}-\frac{26m}{3}+13$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-m\right)^{2}.

$$9-6m+m^{2}+\left(-\frac{2}{3}m+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{2}{3}m+2\right)^{2}.

$$9-6m+m^{2}+\frac{4}{9}m^{2}-\frac{8}{3}m+4$$

Combine m^{2} and \frac{4}{9}m^{2} to get \frac{13}{9}m^{2}.

$$9-6m+\frac{13}{9}m^{2}-\frac{8}{3}m+4$$

Combine -6m and -\frac{8}{3}m to get -\frac{26}{3}m.

$$9-\frac{26}{3}m+\frac{13}{9}m^{2}+4$$

Add 9 and 4 to get 13.

$$13-\frac{26}{3}m+\frac{13}{9}m^{2}$$