Solve (-7a^3+8a^2-3a-3)+(4a^3+6a+3)

$$( - 7 a ^ { 3 } + 8 a ^ { 2 } - 3 a - 3 ) + ( 4 a ^ { 3 } + 6 a + 3 )$$

Evaluate

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

Solution Steps

Combine -7a^{3} and 4a^{3} to get -3a^{3}.

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

Combine -3a and 6a to get 3a.

$$-3a^{3}+8a^{2}+3a-3+3$$

Add -3 and 3 to get 0.

$$-3a^{3}+8a^{2}+3a$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$-3a^{3}+8a^{2}+3a$$

Factor out a.

$$a\left(-3a^{2}+8a+3\right)$$

Consider -3a^{2}+8a+3. Factor the expression by grouping. First, the expression needs to be rewritten as -3a^{2}+pa+qa+3. To find p and q, set up a system to be solved.

$$p+q=8 pq=-3\times 3=-9$$

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 -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$p=9 q=-1$$

Rewrite -3a^{2}+8a+3 as \left(-3a^{2}+9a\right)+\left(-a+3\right).

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

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

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

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

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

Rewrite the complete factored expression.

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