Solve 18x^2+15x-18

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

Factor

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

Solution Steps

Factor out 3.

$$3\left(6x^{2}+5x-6\right)$$

Consider 6x^{2}+5x-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.

$$a+b=5 ab=6\left(-6\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the 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.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite 6x^{2}+5x-6 as \left(6x^{2}-4x\right)+\left(9x-6\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square 15.

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

Multiply -4 times 18.

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

Multiply -72 times -18.

$$x=\frac{-15±\sqrt{225+1296}}{2\times 18}$$

Add 225 to 1296.

$$x=\frac{-15±\sqrt{1521}}{2\times 18}$$

Take the square root of 1521.

$$x=\frac{-15±39}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-15±39}{36}$$

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

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

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±39}{36} when ± is minus. Subtract 39 from -15.

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

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

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

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{3}{2} for x_{2}.

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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

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

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

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

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

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

Multiply 3 times 2.

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

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

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

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

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

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

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

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 = -1

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

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

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

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

$$-u^2 = -1-\frac{25}{144} = -\frac{169}{144}$$

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

$$u^2 = \frac{169}{144} u = \pm\sqrt{\frac{169}{144}} = \pm \frac{13}{12}$$

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

$$r =-\frac{5}{12} - \frac{13}{12} = -1.500 s = -\frac{5}{12} + \frac{13}{12} = 0.667$$

Evaluate

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