Solve 1/8-1/4x+1/6x^2-1/27x^3=

$$\frac { 1 } { 8 } - \frac { 1 } { 4 } x + \frac { 1 } { 6 } x ^ { 2 } - \frac { 1 } { 27 } x ^ { 3 } =$$

Factor

$$\frac{\left(3-2x\right)^{3}}{216}$$

Solution Steps

Factor out \frac{1}{216}.

$$\frac{27-54x+36x^{2}-8x^{3}}{216}$$

Consider 27-54x+36x^{2}-8x^{3}. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient -8. One such root is \frac{3}{2}. Factor the polynomial by dividing it by 2x-3.

$$\left(2x-3\right)\left(-4x^{2}+12x-9\right)$$

Consider -4x^{2}+12x-9. Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=12 ab=-4\left(-9\right)=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$a=6 b=6$$

Rewrite -4x^{2}+12x-9 as \left(-4x^{2}+6x\right)+\left(6x-9\right).

$$\left(-4x^{2}+6x\right)+\left(6x-9\right)$$

Factor out -2x in the first and 3 in the second group.

$$-2x\left(2x-3\right)+3\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-2x+3\right)$$

Rewrite the complete factored expression.

$$\frac{\left(2x-3\right)^{2}\left(-2x+3\right)}{216}$$

Evaluate

$$\frac{\left(3-2x\right)^{3}}{216}$$