Solve 7w^3+10w^2-8w

$$7 w ^ { 3 } + 10 w ^ { 2 } - 8 w$$

Factor

$$w\left(7w-4\right)\left(w+2\right)$$

Solution Steps

Factor out w.

$$w\left(7w^{2}+10w-8\right)$$

Consider 7w^{2}+10w-8. Factor the expression by grouping. First, the expression needs to be rewritten as 7w^{2}+aw+bw-8. To find a and b, set up a system to be solved.

$$a+b=10 ab=7\left(-8\right)=-56$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -56.

$$-1,56 -2,28 -4,14 -7,8$$

Calculate the sum for each pair.

$$-1+56=55 -2+28=26 -4+14=10 -7+8=1$$

The solution is the pair that gives sum 10.

$$a=-4 b=14$$

Rewrite 7w^{2}+10w-8 as \left(7w^{2}-4w\right)+\left(14w-8\right).

$$\left(7w^{2}-4w\right)+\left(14w-8\right)$$

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

$$w\left(7w-4\right)+2\left(7w-4\right)$$

Factor out common term 7w-4 by using distributive property.

$$\left(7w-4\right)\left(w+2\right)$$

Rewrite the complete factored expression.

$$w\left(7w-4\right)\left(w+2\right)$$

Evaluate

$$w\left(7w-4\right)\left(w+2\right)$$