Solve 81z^3+36z^2+3z

$$81 z ^ { 3 } + 36 z ^ { 2 } + 3 z$$

Factor

$$3z\left(3z+1\right)\left(9z+1\right)$$

Solution Steps

Factor out 3.

$$3\left(27z^{3}+12z^{2}+z\right)$$

Consider 27z^{3}+12z^{2}+z. Factor out z.

$$z\left(27z^{2}+12z+1\right)$$

Consider 27z^{2}+12z+1. Factor the expression by grouping. First, the expression needs to be rewritten as 27z^{2}+az+bz+1. To find a and b, set up a system to be solved.

$$a+b=12 ab=27\times 1=27$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 27.

$$1,27 3,9$$

Calculate the sum for each pair.

$$1+27=28 3+9=12$$

The solution is the pair that gives sum 12.

$$a=3 b=9$$

Rewrite 27z^{2}+12z+1 as \left(27z^{2}+3z\right)+\left(9z+1\right).

$$\left(27z^{2}+3z\right)+\left(9z+1\right)$$

Factor out 3z in 27z^{2}+3z.

$$3z\left(9z+1\right)+9z+1$$

Factor out common term 9z+1 by using distributive property.

$$\left(9z+1\right)\left(3z+1\right)$$

Rewrite the complete factored expression.

$$3z\left(9z+1\right)\left(3z+1\right)$$

Evaluate

$$3z\left(3z+1\right)\left(9z+1\right)$$