Solve 2a^2-2/3a-4/9)*(-5

$$2 a ^ { 2 } - \frac { 2 } { 3 } a - \frac { 4 } { 9 } ) \cdot ( - 5$$

Factor

$$\frac{2\left(3a-2\right)\left(3a+1\right)}{9}$$

Solution Steps

Factor out \frac{2}{9}.

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

Consider 9a^{2}-3a-2. Factor the expression by grouping. First, the expression needs to be rewritten as 9a^{2}+pa+qa-2. To find p and q, set up a system to be solved.

$$p+q=-3 pq=9\left(-2\right)=-18$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$p=-6 q=3$$

Rewrite 9a^{2}-3a-2 as \left(9a^{2}-6a\right)+\left(3a-2\right).

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

Factor out 3a in 9a^{2}-6a.

$$3a\left(3a-2\right)+3a-2$$

Factor out common term 3a-2 by using distributive property.

$$\left(3a-2\right)\left(3a+1\right)$$

Rewrite the complete factored expression.

$$\frac{2\left(3a-2\right)\left(3a+1\right)}{9}$$

Evaluate

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