Solve 18x^2-27x-5

$$18 x ^ { 2 } - 27 x - 5$$

Factor

$$\left(3x-5\right)\left(6x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-27 ab=18\left(-5\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -27.

$$a=-30 b=3$$

Rewrite 18x^{2}-27x-5 as \left(18x^{2}-30x\right)+\left(3x-5\right).

$$\left(18x^{2}-30x\right)+\left(3x-5\right)$$

Factor out 6x in 18x^{2}-30x.

$$6x\left(3x-5\right)+3x-5$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(6x+1\right)$$

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.

$$18x^{2}-27x-5=0$$

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.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 18\left(-5\right)}}{2\times 18}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 18\left(-5\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-27\right)±\sqrt{729-72\left(-5\right)}}{2\times 18}$$

Multiply -72 times -5.

$$x=\frac{-\left(-27\right)±\sqrt{729+360}}{2\times 18}$$

Add 729 to 360.

$$x=\frac{-\left(-27\right)±\sqrt{1089}}{2\times 18}$$

Take the square root of 1089.

$$x=\frac{-\left(-27\right)±33}{2\times 18}$$

The opposite of -27 is 27.

$$x=\frac{27±33}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{27±33}{36}$$

Now solve the equation x=\frac{27±33}{36} when ± is plus. Add 27 to 33.

$$x=\frac{60}{36}$$

Reduce the fraction \frac{60}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{27±33}{36} when ± is minus. Subtract 33 from 27.

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

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{5}{3} for x_{1} and -\frac{1}{6} for x_{2}.

$$18x^{2}-27x-5=18\left(x-\frac{5}{3}\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$18x^{2}-27x-5=18\left(x-\frac{5}{3}\right)\left(x+\frac{1}{6}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-27x-5=18\times \frac{3x-5}{3}\left(x+\frac{1}{6}\right)$$

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

$$18x^{2}-27x-5=18\times \frac{3x-5}{3}\times \frac{6x+1}{6}$$

Multiply \frac{3x-5}{3} times \frac{6x+1}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-27x-5=18\times \frac{\left(3x-5\right)\left(6x+1\right)}{3\times 6}$$

Multiply 3 times 6.

$$18x^{2}-27x-5=18\times \frac{\left(3x-5\right)\left(6x+1\right)}{18}$$

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

$$18x^{2}-27x-5=\left(3x-5\right)\left(6x+1\right)$$

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

$$x ^ 2 -\frac{3}{2}x -\frac{5}{18} = 0$$

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

$$r + s = \frac{3}{2} rs = -\frac{5}{18}$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{18}

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -\frac{5}{18}$$

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

$$\frac{9}{16} - u^2 = -\frac{5}{18}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{5}{18}-\frac{9}{16} = -\frac{121}{144}$$

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.

$$r =\frac{3}{4} - \frac{11}{12} = -0.167 s = \frac{3}{4} + \frac{11}{12} = 1.667$$

Evaluate

$$\left(3x-5\right)\left(6x+1\right)$$