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

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

Factor

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

Solution Steps

Factor out b^{3}.

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

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

$$p+q=-10 pq=1\times 16=16$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q 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.

$$p=-8 q=-2$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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