Solve 18x^2-15x-12

$$18 x ^ { 2 } - 15 x - 12$$

Factor

$$3\left(3x-4\right)\left(2x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(6x^{2}-5x-4\right)$$

Consider 6x^{2}-5x-4. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\left(-4\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=-8 b=3$$

Rewrite 6x^{2}-5x-4 as \left(6x^{2}-8x\right)+\left(3x-4\right).

$$\left(6x^{2}-8x\right)+\left(3x-4\right)$$

Factor out 2x in 6x^{2}-8x.

$$2x\left(3x-4\right)+3x-4$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(2x+1\right)$$

Rewrite the complete factored expression.

$$3\left(3x-4\right)\left(2x+1\right)$$

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.

$$18x^{2}-15x-12=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 18\left(-12\right)}}{2\times 18}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 18\left(-12\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-15\right)±\sqrt{225-72\left(-12\right)}}{2\times 18}$$

Multiply -72 times -12.

$$x=\frac{-\left(-15\right)±\sqrt{225+864}}{2\times 18}$$

Add 225 to 864.

$$x=\frac{-\left(-15\right)±\sqrt{1089}}{2\times 18}$$

Take the square root of 1089.

$$x=\frac{-\left(-15\right)±33}{2\times 18}$$

The opposite of -15 is 15.

$$x=\frac{15±33}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{15±33}{36}$$

Now solve the equation x=\frac{15±33}{36} when ± is plus. Add 15 to 33.

$$x=\frac{48}{36}$$

Reduce the fraction \frac{48}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{15±33}{36} when ± is minus. Subtract 33 from 15.

$$x=-\frac{18}{36}$$

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

$$x=-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{1}{2} for x_{2}.

$$18x^{2}-15x-12=18\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$18x^{2}-15x-12=18\left(x-\frac{4}{3}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-15x-12=18\times \frac{3x-4}{3}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-15x-12=18\times \frac{3x-4}{3}\times \frac{2x+1}{2}$$

Multiply \frac{3x-4}{3} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-15x-12=18\times \frac{\left(3x-4\right)\left(2x+1\right)}{3\times 2}$$

Multiply 3 times 2.

$$18x^{2}-15x-12=18\times \frac{\left(3x-4\right)\left(2x+1\right)}{6}$$

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

$$18x^{2}-15x-12=3\left(3x-4\right)\left(2x+1\right)$$

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

$$x ^ 2 -\frac{5}{6}x -\frac{2}{3} = 0$$

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

$$r + s = \frac{5}{6} rs = -\frac{2}{3}$$

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{2}{3}

$$(\frac{5}{12} - u) (\frac{5}{12} + u) = -\frac{2}{3}$$

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

$$\frac{25}{144} - u^2 = -\frac{2}{3}$$

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

$$-u^2 = -\frac{2}{3}-\frac{25}{144} = -\frac{121}{144}$$

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

$$u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}$$

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

$$r =\frac{5}{12} - \frac{11}{12} = -0.500 s = \frac{5}{12} + \frac{11}{12} = 1.333$$

Evaluate

$$18x^{2}-15x-12$$