Solve w^5-13w^4+42w^3

$$w ^ { 5 } - 13 w ^ { 4 } + 42 w ^ { 3 }$$

Factor

$$\left(w-7\right)\left(w-6\right)w^{3}$$

Solution Steps

Factor out w^{3}.

$$w^{3}\left(w^{2}-13w+42\right)$$

Consider w^{2}-13w+42. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw+42. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 42=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite w^{2}-13w+42 as \left(w^{2}-7w\right)+\left(-6w+42\right).

$$\left(w^{2}-7w\right)+\left(-6w+42\right)$$

Factor out w in the first and -6 in the second group.

$$w\left(w-7\right)-6\left(w-7\right)$$

Factor out common term w-7 by using distributive property.

$$\left(w-7\right)\left(w-6\right)$$

Rewrite the complete factored expression.

$$w^{3}\left(w-7\right)\left(w-6\right)$$

Evaluate

$$\left(w-7\right)\left(w-6\right)w^{3}$$