Solve 0-5k^3+40k^2-80k

$$0 - 5 k ^ { 3 } + 40 k ^ { 2 } - 80 k$$

Factor

$$-5k\left(4-k\right)^{2}$$

Solution Steps

Factor out 5.

$$5\left(-k^{3}+8k^{2}-16k\right)$$

Consider -k^{3}+8k^{2}-16k. Factor out k.

$$k\left(-k^{2}+8k-16\right)$$

Consider -k^{2}+8k-16. Factor the expression by grouping. First, the expression needs to be rewritten as -k^{2}+ak+bk-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite -k^{2}+8k-16 as \left(-k^{2}+4k\right)+\left(4k-16\right).

$$\left(-k^{2}+4k\right)+\left(4k-16\right)$$

Factor out -k in the first and 4 in the second group.

$$-k\left(k-4\right)+4\left(k-4\right)$$

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

$$\left(k-4\right)\left(-k+4\right)$$

Rewrite the complete factored expression.

$$5k\left(k-4\right)\left(-k+4\right)$$

Evaluate

$$-5k\left(4-k\right)^{2}$$

Solution Steps

Anything plus zero gives itself.

$$-5k^{3}+40k^{2}-80k$$