Solve 30x^2+33x-18

$$30 x ^ { 2 } + 33 x - 18$$

Factor

$$3\left(5x-2\right)\left(2x+3\right)$$

Solution Steps

Factor out 3.

$$3\left(10x^{2}+11x-6\right)$$

Consider 10x^{2}+11x-6. Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=11 ab=10\left(-6\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=-4 b=15$$

Rewrite 10x^{2}+11x-6 as \left(10x^{2}-4x\right)+\left(15x-6\right).

$$\left(10x^{2}-4x\right)+\left(15x-6\right)$$

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

$$2x\left(5x-2\right)+3\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(2x+3\right)$$

Rewrite the complete factored expression.

$$3\left(5x-2\right)\left(2x+3\right)$$

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.

$$30x^{2}+33x-18=0$$

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.

$$x=\frac{-33±\sqrt{33^{2}-4\times 30\left(-18\right)}}{2\times 30}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\times 30\left(-18\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-33±\sqrt{1089-120\left(-18\right)}}{2\times 30}$$

Multiply -120 times -18.

$$x=\frac{-33±\sqrt{1089+2160}}{2\times 30}$$

Add 1089 to 2160.

$$x=\frac{-33±\sqrt{3249}}{2\times 30}$$

Take the square root of 3249.

$$x=\frac{-33±57}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{-33±57}{60}$$

Now solve the equation x=\frac{-33±57}{60} when ± is plus. Add -33 to 57.

$$x=\frac{24}{60}$$

Reduce the fraction \frac{24}{60} to lowest terms by extracting and canceling out 12.

$$x=\frac{2}{5}$$

Now solve the equation x=\frac{-33±57}{60} when ± is minus. Subtract 57 from -33.

$$x=-\frac{90}{60}$$

Reduce the fraction \frac{-90}{60} to lowest terms by extracting and canceling out 30.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{3}{2} for x_{2}.

$$30x^{2}+33x-18=30\left(x-\frac{2}{5}\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$30x^{2}+33x-18=30\left(x-\frac{2}{5}\right)\left(x+\frac{3}{2}\right)$$

Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+33x-18=30\times \frac{5x-2}{5}\left(x+\frac{3}{2}\right)$$

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

$$30x^{2}+33x-18=30\times \frac{5x-2}{5}\times \frac{2x+3}{2}$$

Multiply \frac{5x-2}{5} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+33x-18=30\times \frac{\left(5x-2\right)\left(2x+3\right)}{5\times 2}$$

Multiply 5 times 2.

$$30x^{2}+33x-18=30\times \frac{\left(5x-2\right)\left(2x+3\right)}{10}$$

Cancel out 10, the greatest common factor in 30 and 10.

$$30x^{2}+33x-18=3\left(5x-2\right)\left(2x+3\right)$$

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 30

$$x ^ 2 +\frac{11}{10}x -\frac{3}{5} = 0$$

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

$$r + s = -\frac{11}{10} rs = -\frac{3}{5}$$

Two numbers r and s sum up to -\frac{11}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{10} = -\frac{11}{20}. 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>

$$r = -\frac{11}{20} - u s = -\frac{11}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}

$$(-\frac{11}{20} - u) (-\frac{11}{20} + u) = -\frac{3}{5}$$

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

$$\frac{121}{400} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{121}{400} on both sides

$$-u^2 = -\frac{3}{5}-\frac{121}{400} = -\frac{361}{400}$$

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

$$u^2 = \frac{361}{400} u = \pm\sqrt{\frac{361}{400}} = \pm \frac{19}{20}$$

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

$$r =-\frac{11}{20} - \frac{19}{20} = -1.500 s = -\frac{11}{20} + \frac{19}{20} = 0.400$$

Evaluate

$$30x^{2}+33x-18$$