Solve (a)^3+a^2-6a):(a-2)

$$( a ) ^ { 3 } + a ^ { 2 } - 6 a ) : ( a - 2 )$$

Factor

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

Solution Steps

Factor out a.

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

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

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

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$p=-2 q=3$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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