Solve q-3q^2-18q^3

$$q-3 { q }^{ 2 } -18 { q }^{ 3 }$$

Factor

$$q\left(1-6q\right)\left(3q+1\right)$$

Solution Steps

Factor out q.

$$q\left(1-3q-18q^{2}\right)$$

Consider 1-3q-18q^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-18q^{2}-3q+1$$

Factor the expression by grouping. First, the expression needs to be rewritten as -18q^{2}+aq+bq+1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-18=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=3 b=-6$$

Rewrite -18q^{2}-3q+1 as \left(-18q^{2}+3q\right)+\left(-6q+1\right).

$$\left(-18q^{2}+3q\right)+\left(-6q+1\right)$$

Factor out 3q in -18q^{2}+3q.

$$3q\left(-6q+1\right)-6q+1$$

Factor out common term -6q+1 by using distributive property.

$$\left(-6q+1\right)\left(3q+1\right)$$

Rewrite the complete factored expression.

$$q\left(-6q+1\right)\left(3q+1\right)$$

Evaluate

$$q\left(1-6q\right)\left(3q+1\right)$$