Solve a^2*0.2+2*a*(1-a)*-0.2+left(1-aright)^2*03

$${ a }^{ 2 } \times 0.2+2 \times a \times (1-a) \times -0.2+ { \left(1-a \right) }^{ 2 } \times 03$$

Evaluate

$$a\times \frac{3a-2}{5}$$

Solution Steps

Multiply 2 and -0.2 to get -0.4.

$$a^{2}\times 0.2-0.4a\left(1-a\right)+\left(1-a\right)^{2}\times 0\times 3$$

Use the distributive property to multiply -0.4a by 1-a.

$$a^{2}\times 0.2-0.4a+0.4a^{2}+\left(1-a\right)^{2}\times 0\times 3$$

Combine a^{2}\times 0.2 and 0.4a^{2} to get 0.6a^{2}.

$$0.6a^{2}-0.4a+\left(1-a\right)^{2}\times 0\times 3$$

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

$$0.6a^{2}-0.4a+\left(1-2a+a^{2}\right)\times 0\times 3$$

Multiply 0 and 3 to get 0.

$$0.6a^{2}-0.4a+\left(1-2a+a^{2}\right)\times 0$$

Anything times zero gives zero.

$$0.6a^{2}-0.4a+0$$

Anything plus zero gives itself.

$$0.6a^{2}-0.4a$$

Expand

$$\frac{3a^{2}-2a}{5}$$

Solution Steps

Multiply 2 and -0.2 to get -0.4.

$$a^{2}\times 0.2-0.4a\left(1-a\right)+\left(1-a\right)^{2}\times 0\times 3$$

Use the distributive property to multiply -0.4a by 1-a.

$$a^{2}\times 0.2-0.4a+0.4a^{2}+\left(1-a\right)^{2}\times 0\times 3$$

Combine a^{2}\times 0.2 and 0.4a^{2} to get 0.6a^{2}.

$$0.6a^{2}-0.4a+\left(1-a\right)^{2}\times 0\times 3$$

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

$$0.6a^{2}-0.4a+\left(1-2a+a^{2}\right)\times 0\times 3$$

Multiply 0 and 3 to get 0.

$$0.6a^{2}-0.4a+\left(1-2a+a^{2}\right)\times 0$$

Anything times zero gives zero.

$$0.6a^{2}-0.4a+0$$

Anything plus zero gives itself.

$$0.6a^{2}-0.4a$$