Solve 18x^2-39x+18

$$18 { x }^{ 2 } -39x+18$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(6x^{2}-13x+6\right)$$

Consider 6x^{2}-13x+6. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=6\times 6=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 -13.

$$a=-9 b=-4$$

Rewrite 6x^{2}-13x+6 as \left(6x^{2}-9x\right)+\left(-4x+6\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$18x^{2}-39x+18=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 18\times 18}}{2\times 18}$$

Square -39.

$$x=\frac{-\left(-39\right)±\sqrt{1521-4\times 18\times 18}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-39\right)±\sqrt{1521-72\times 18}}{2\times 18}$$

Multiply -72 times 18.

$$x=\frac{-\left(-39\right)±\sqrt{1521-1296}}{2\times 18}$$

Add 1521 to -1296.

$$x=\frac{-\left(-39\right)±\sqrt{225}}{2\times 18}$$

Take the square root of 225.

$$x=\frac{-\left(-39\right)±15}{2\times 18}$$

The opposite of -39 is 39.

$$x=\frac{39±15}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{39±15}{36}$$

Now solve the equation x=\frac{39±15}{36} when ± is plus. Add 39 to 15.

$$x=\frac{54}{36}$$

Reduce the fraction \frac{54}{36} to lowest terms by extracting and canceling out 18.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{39±15}{36} when ± is minus. Subtract 15 from 39.

$$x=\frac{24}{36}$$

Reduce the fraction \frac{24}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{2}{3} for x_{2}.

$$18x^{2}-39x+18=18\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-39x+18=18\times \frac{2x-3}{2}\left(x-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-39x+18=18\times \frac{2x-3}{2}\times \frac{3x-2}{3}$$

Multiply \frac{2x-3}{2} times \frac{3x-2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-39x+18=18\times \frac{\left(2x-3\right)\left(3x-2\right)}{2\times 3}$$

Multiply 2 times 3.

$$18x^{2}-39x+18=18\times \frac{\left(2x-3\right)\left(3x-2\right)}{6}$$

Cancel out 6, the greatest common factor in 18 and 6.

$$18x^{2}-39x+18=3\left(2x-3\right)\left(3x-2\right)$$

Evaluate

$$18x^{2}-39x+18$$