Solve 40c^3+22c^2-6c

$$40 c ^ { 3 } + 22 c ^ { 2 } - 6 c$$

Factor

$$2c\left(5c-1\right)\left(4c+3\right)$$

Solution Steps

Factor out 2.

$$2\left(20c^{3}+11c^{2}-3c\right)$$

Consider 20c^{3}+11c^{2}-3c. Factor out c.

$$c\left(20c^{2}+11c-3\right)$$

Consider 20c^{2}+11c-3. Factor the expression by grouping. First, the expression needs to be rewritten as 20c^{2}+ac+bc-3. To find a and b, set up a system to be solved.

$$a+b=11 ab=20\left(-3\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=-4 b=15$$

Rewrite 20c^{2}+11c-3 as \left(20c^{2}-4c\right)+\left(15c-3\right).

$$\left(20c^{2}-4c\right)+\left(15c-3\right)$$

Factor out 4c in the first and 3 in the second group.

$$4c\left(5c-1\right)+3\left(5c-1\right)$$

Factor out common term 5c-1 by using distributive property.

$$\left(5c-1\right)\left(4c+3\right)$$

Rewrite the complete factored expression.

$$2c\left(5c-1\right)\left(4c+3\right)$$

Evaluate

$$2c\left(5c-1\right)\left(4c+3\right)$$