Solve 18x^2-15x+2

$$18 x ^ { 2 } - 15 x + 2$$

Factor

$$\left(3x-2\right)\left(6x-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-15 ab=18\times 2=36$$

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 -15.

$$a=-12 b=-3$$

Rewrite 18x^{2}-15x+2 as \left(18x^{2}-12x\right)+\left(-3x+2\right).

$$\left(18x^{2}-12x\right)+\left(-3x+2\right)$$

Factor out 6x in the first and -1 in the second group.

$$6x\left(3x-2\right)-\left(3x-2\right)$$

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

$$\left(3x-2\right)\left(6x-1\right)$$

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.

$$18x^{2}-15x+2=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 18\times 2}}{2\times 18}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 18\times 2}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-15\right)±\sqrt{225-72\times 2}}{2\times 18}$$

Multiply -72 times 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 18}$$

Add 225 to -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 18}$$

Take the square root of 81.

$$x=\frac{-\left(-15\right)±9}{2\times 18}$$

The opposite of -15 is 15.

$$x=\frac{15±9}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{15±9}{36}$$

Now solve the equation x=\frac{15±9}{36} when ± is plus. Add 15 to 9.

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

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

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

Now solve the equation x=\frac{15±9}{36} when ± is minus. Subtract 9 from 15.

$$x=\frac{6}{36}$$

Reduce the fraction \frac{6}{36} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and \frac{1}{6} for x_{2}.

$$18x^{2}-15x+2=18\left(x-\frac{2}{3}\right)\left(x-\frac{1}{6}\right)$$

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

$$18x^{2}-15x+2=18\times \frac{3x-2}{3}\left(x-\frac{1}{6}\right)$$

Subtract \frac{1}{6} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-15x+2=18\times \frac{3x-2}{3}\times \frac{6x-1}{6}$$

Multiply \frac{3x-2}{3} times \frac{6x-1}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-15x+2=18\times \frac{\left(3x-2\right)\left(6x-1\right)}{3\times 6}$$

Multiply 3 times 6.

$$18x^{2}-15x+2=18\times \frac{\left(3x-2\right)\left(6x-1\right)}{18}$$

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

$$18x^{2}-15x+2=\left(3x-2\right)\left(6x-1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 18

$$x ^ 2 -\frac{5}{6}x +\frac{1}{9} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{5}{6} rs = \frac{1}{9}$$

Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{12} - u s = \frac{5}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{9}

$$(\frac{5}{12} - u) (\frac{5}{12} + u) = \frac{1}{9}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{144} - u^2 = \frac{1}{9}$$

Simplify the expression by subtracting \frac{25}{144} on both sides

$$-u^2 = \frac{1}{9}-\frac{25}{144} = -\frac{1}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{5}{12} - \frac{1}{4} = 0.167 s = \frac{5}{12} + \frac{1}{4} = 0.667$$

Evaluate

$$\left(3x-2\right)\left(6x-1\right)$$