Solve 30x^2-9x-12

$$30 x ^ { 2 } - 9 x - 12$$

Factor

$$3\left(5x-4\right)\left(2x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(10x^{2}-3x-4\right)$$

Consider 10x^{2}-3x-4. Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=10\left(-4\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite 10x^{2}-3x-4 as \left(10x^{2}-8x\right)+\left(5x-4\right).

$$\left(10x^{2}-8x\right)+\left(5x-4\right)$$

Factor out 2x in 10x^{2}-8x.

$$2x\left(5x-4\right)+5x-4$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(2x+1\right)$$

Rewrite the complete factored expression.

$$3\left(5x-4\right)\left(2x+1\right)$$

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.

$$30x^{2}-9x-12=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 30\left(-12\right)}}{2\times 30}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 30\left(-12\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-\left(-9\right)±\sqrt{81-120\left(-12\right)}}{2\times 30}$$

Multiply -120 times -12.

$$x=\frac{-\left(-9\right)±\sqrt{81+1440}}{2\times 30}$$

Add 81 to 1440.

$$x=\frac{-\left(-9\right)±\sqrt{1521}}{2\times 30}$$

Take the square root of 1521.

$$x=\frac{-\left(-9\right)±39}{2\times 30}$$

The opposite of -9 is 9.

$$x=\frac{9±39}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{9±39}{60}$$

Now solve the equation x=\frac{9±39}{60} when ± is plus. Add 9 to 39.

$$x=\frac{48}{60}$$

Reduce the fraction \frac{48}{60} to lowest terms by extracting and canceling out 12.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{9±39}{60} when ± is minus. Subtract 39 from 9.

$$x=-\frac{30}{60}$$

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{5} for x_{1} and -\frac{1}{2} for x_{2}.

$$30x^{2}-9x-12=30\left(x-\frac{4}{5}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$30x^{2}-9x-12=30\left(x-\frac{4}{5}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{4}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}-9x-12=30\times \frac{5x-4}{5}\left(x+\frac{1}{2}\right)$$

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

$$30x^{2}-9x-12=30\times \frac{5x-4}{5}\times \frac{2x+1}{2}$$

Multiply \frac{5x-4}{5} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}-9x-12=30\times \frac{\left(5x-4\right)\left(2x+1\right)}{5\times 2}$$

Multiply 5 times 2.

$$30x^{2}-9x-12=30\times \frac{\left(5x-4\right)\left(2x+1\right)}{10}$$

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

$$30x^{2}-9x-12=3\left(5x-4\right)\left(2x+1\right)$$

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 30

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

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

$$r + s = \frac{3}{10} rs = -\frac{2}{5}$$

Two numbers r and s sum up to \frac{3}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{10} = \frac{3}{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>

$$r = \frac{3}{20} - u s = \frac{3}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}

$$(\frac{3}{20} - u) (\frac{3}{20} + u) = -\frac{2}{5}$$

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

$$\frac{9}{400} - u^2 = -\frac{2}{5}$$

Simplify the expression by subtracting \frac{9}{400} on both sides

$$-u^2 = -\frac{2}{5}-\frac{9}{400} = -\frac{169}{400}$$

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

$$u^2 = \frac{169}{400} u = \pm\sqrt{\frac{169}{400}} = \pm \frac{13}{20}$$

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

$$r =\frac{3}{20} - \frac{13}{20} = -0.500 s = \frac{3}{20} + \frac{13}{20} = 0.800$$

Evaluate

$$30x^{2}-9x-12$$