Solve w^5-10w^4+16w^3

$$w ^ { 5 } - 10 w ^ { 4 } + 16 w ^ { 3 }$$

Factor

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

Solution Steps

Factor out w^{3}.

$$w^{3}\left(w^{2}-10w+16\right)$$

Consider w^{2}-10w+16. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw+16. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 16=16$$

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 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 -10.

$$a=-8 b=-2$$

Rewrite w^{2}-10w+16 as \left(w^{2}-8w\right)+\left(-2w+16\right).

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

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

$$w\left(w-8\right)-2\left(w-8\right)$$

Factor out common term w-8 by using distributive property.

$$\left(w-8\right)\left(w-2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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