Solve 1/3x^2-2x-9

$$\frac{ 1 }{ 3 } { x }^{ 2 } -2x-9$$

Factor

$$\frac{\left(x-9\right)\left(x+3\right)}{3}$$

Solution Steps

Factor out \frac{1}{3}.

$$\frac{x^{2}-6x-27}{3}$$

Consider x^{2}-6x-27. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-27\right)=-27$$

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 -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -6.

$$a=-9 b=3$$

Rewrite x^{2}-6x-27 as \left(x^{2}-9x\right)+\left(3x-27\right).

$$\left(x^{2}-9x\right)+\left(3x-27\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-9\right)+3\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$\frac{\left(x-9\right)\left(x+3\right)}{3}$$

Evaluate

$$\frac{\left(x-9\right)\left(x+3\right)}{3}$$