Solve -2a^3+12a^2-16a

$$- 2 a ^ { 3 } + 12 a ^ { 2 } - 16 a$$

Factor

$$2a\left(2-a\right)\left(a-4\right)$$

Solution Steps

Factor out 2.

$$2\left(-a^{3}+6a^{2}-8a\right)$$

Consider -a^{3}+6a^{2}-8a. Factor out a.

$$a\left(-a^{2}+6a-8\right)$$

Consider -a^{2}+6a-8. Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-8. To find p and q, set up a system to be solved.

$$p+q=6 pq=-\left(-8\right)=8$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$p=4 q=2$$

Rewrite -a^{2}+6a-8 as \left(-a^{2}+4a\right)+\left(2a-8\right).

$$\left(-a^{2}+4a\right)+\left(2a-8\right)$$

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

$$-a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(-a+2\right)$$

Rewrite the complete factored expression.

$$2a\left(a-4\right)\left(-a+2\right)$$

Evaluate

$$2a\left(2-a\right)\left(a-4\right)$$