Solve 27x^2-30x-25

$$27 x ^ { 2 } - 30 x - 25$$

Factor

$$\left(3x-5\right)\left(9x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 27x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=-30 ab=27\left(-25\right)=-675$$

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 -675.

$$1,-675 3,-225 5,-135 9,-75 15,-45 25,-27$$

Calculate the sum for each pair.

$$1-675=-674 3-225=-222 5-135=-130 9-75=-66 15-45=-30 25-27=-2$$

The solution is the pair that gives sum -30.

$$a=-45 b=15$$

Rewrite 27x^{2}-30x-25 as \left(27x^{2}-45x\right)+\left(15x-25\right).

$$\left(27x^{2}-45x\right)+\left(15x-25\right)$$

Factor out 9x in the first and 5 in the second group.

$$9x\left(3x-5\right)+5\left(3x-5\right)$$

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

$$\left(3x-5\right)\left(9x+5\right)$$

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.

$$27x^{2}-30x-25=0$$

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.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 27\left(-25\right)}}{2\times 27}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 27\left(-25\right)}}{2\times 27}$$

Multiply -4 times 27.

$$x=\frac{-\left(-30\right)±\sqrt{900-108\left(-25\right)}}{2\times 27}$$

Multiply -108 times -25.

$$x=\frac{-\left(-30\right)±\sqrt{900+2700}}{2\times 27}$$

Add 900 to 2700.

$$x=\frac{-\left(-30\right)±\sqrt{3600}}{2\times 27}$$

Take the square root of 3600.

$$x=\frac{-\left(-30\right)±60}{2\times 27}$$

The opposite of -30 is 30.

$$x=\frac{30±60}{2\times 27}$$

Multiply 2 times 27.

$$x=\frac{30±60}{54}$$

Now solve the equation x=\frac{30±60}{54} when ± is plus. Add 30 to 60.

$$x=\frac{90}{54}$$

Reduce the fraction \frac{90}{54} to lowest terms by extracting and canceling out 18.

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

Now solve the equation x=\frac{30±60}{54} when ± is minus. Subtract 60 from 30.

$$x=-\frac{30}{54}$$

Reduce the fraction \frac{-30}{54} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{9}$$

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{5}{9} for x_{2}.

$$27x^{2}-30x-25=27\left(x-\frac{5}{3}\right)\left(x-\left(-\frac{5}{9}\right)\right)$$

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

$$27x^{2}-30x-25=27\left(x-\frac{5}{3}\right)\left(x+\frac{5}{9}\right)$$

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

$$27x^{2}-30x-25=27\times \frac{3x-5}{3}\left(x+\frac{5}{9}\right)$$

Add \frac{5}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$27x^{2}-30x-25=27\times \frac{3x-5}{3}\times \frac{9x+5}{9}$$

Multiply \frac{3x-5}{3} times \frac{9x+5}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$27x^{2}-30x-25=27\times \frac{\left(3x-5\right)\left(9x+5\right)}{3\times 9}$$

Multiply 3 times 9.

$$27x^{2}-30x-25=27\times \frac{\left(3x-5\right)\left(9x+5\right)}{27}$$

Cancel out 27, the greatest common factor in 27 and 27.

$$27x^{2}-30x-25=\left(3x-5\right)\left(9x+5\right)$$

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 27

$$x ^ 2 -\frac{10}{9}x -\frac{25}{27} = 0$$

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

$$r + s = \frac{10}{9} rs = -\frac{25}{27}$$

Two numbers r and s sum up to \frac{10}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{9} = \frac{5}{9}. 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>

$$r = \frac{5}{9} - u s = \frac{5}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{25}{27}

$$(\frac{5}{9} - u) (\frac{5}{9} + u) = -\frac{25}{27}$$

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

$$\frac{25}{81} - u^2 = -\frac{25}{27}$$

Simplify the expression by subtracting \frac{25}{81} on both sides

$$-u^2 = -\frac{25}{27}-\frac{25}{81} = -\frac{100}{81}$$

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

$$u^2 = \frac{100}{81} u = \pm\sqrt{\frac{100}{81}} = \pm \frac{10}{9}$$

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

$$r =\frac{5}{9} - \frac{10}{9} = -0.556 s = \frac{5}{9} + \frac{10}{9} = 1.667$$

Evaluate

$$\left(3x-5\right)\left(9x+5\right)$$