Solve w^3-6w^2+8w

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

Factor

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

Solution Steps

Factor out w.

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

Consider w^{2}-6w+8. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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.

$$a=-4 b=-2$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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