Solve (2a^3-8a^2+3a)+(4a^3+9a^2-5a)

$$( 2 a ^ { 3 } - 8 a ^ { 2 } + 3 a ) + ( 4 a ^ { 3 } + 9 a ^ { 2 } - 5 a )$$

Evaluate

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

Solution Steps

Combine 2a^{3} and 4a^{3} to get 6a^{3}.

$$6a^{3}-8a^{2}+3a+9a^{2}-5a$$

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

$$6a^{3}+a^{2}+3a-5a$$

Combine 3a and -5a to get -2a.

$$6a^{3}+a^{2}-2a$$

Factor

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

Solution Steps

Factor out a.

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

Consider 2a^{2}-8a+3+4a^{2}+9a-5. Multiply and combine like terms.

$$6a^{2}+a-2$$

Consider 6a^{2}+a-2. Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-2. To find p and q, set up a system to be solved.

$$p+q=1 pq=6\left(-2\right)=-12$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$p=-3 q=4$$

Rewrite 6a^{2}+a-2 as \left(6a^{2}-3a\right)+\left(4a-2\right).

$$\left(6a^{2}-3a\right)+\left(4a-2\right)$$

Factor out 3a in the first and 2 in the second group.

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

Factor out common term 2a-1 by using distributive property.

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

Rewrite the complete factored expression.

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