Solve 8c^3+26c^2-24c

$$8 c ^ { 3 } + 26 c ^ { 2 } - 24 c$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(4c^{3}+13c^{2}-12c\right)$$

Consider 4c^{3}+13c^{2}-12c. Factor out c.

$$c\left(4c^{2}+13c-12\right)$$

Consider 4c^{2}+13c-12. Factor the expression by grouping. First, the expression needs to be rewritten as 4c^{2}+ac+bc-12. To find a and b, set up a system to be solved.

$$a+b=13 ab=4\left(-12\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 13.

$$a=-3 b=16$$

Rewrite 4c^{2}+13c-12 as \left(4c^{2}-3c\right)+\left(16c-12\right).

$$\left(4c^{2}-3c\right)+\left(16c-12\right)$$

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

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

Factor out common term 4c-3 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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