Solve 12x^2-30x-108rightarrow

$$12 x ^ { 2 } - 30 x - 108 \rightarrow$$

Factor

$$6\left(2x-9\right)\left(x+2\right)$$

Solution Steps

Factor out 6.

$$6\left(2x^{2}-5x-18\right)$$

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

$$a+b=-5 ab=2\left(-18\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite 2x^{2}-5x-18 as \left(2x^{2}-9x\right)+\left(4x-18\right).

$$\left(2x^{2}-9x\right)+\left(4x-18\right)$$

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

$$x\left(2x-9\right)+2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$6\left(2x-9\right)\left(x+2\right)$$

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.

$$12x^{2}-30x-108=0$$

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.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 12\left(-108\right)}}{2\times 12}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 12\left(-108\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-30\right)±\sqrt{900-48\left(-108\right)}}{2\times 12}$$

Multiply -48 times -108.

$$x=\frac{-\left(-30\right)±\sqrt{900+5184}}{2\times 12}$$

Add 900 to 5184.

$$x=\frac{-\left(-30\right)±\sqrt{6084}}{2\times 12}$$

Take the square root of 6084.

$$x=\frac{-\left(-30\right)±78}{2\times 12}$$

The opposite of -30 is 30.

$$x=\frac{30±78}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{30±78}{24}$$

Now solve the equation x=\frac{30±78}{24} when ± is plus. Add 30 to 78.

$$x=\frac{108}{24}$$

Reduce the fraction \frac{108}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{30±78}{24} when ± is minus. Subtract 78 from 30.

$$x=-\frac{48}{24}$$

Divide -48 by 24.

$$x=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and -2 for x_{2}.

$$12x^{2}-30x-108=12\left(x-\frac{9}{2}\right)\left(x-\left(-2\right)\right)$$

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

$$12x^{2}-30x-108=12\left(x-\frac{9}{2}\right)\left(x+2\right)$$

Subtract \frac{9}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-30x-108=12\times \frac{2x-9}{2}\left(x+2\right)$$

Cancel out 2, the greatest common factor in 12 and 2.

$$12x^{2}-30x-108=6\left(2x-9\right)\left(x+2\right)$$

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 12

$$x ^ 2 -\frac{5}{2}x -9 = 0$$

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

$$r + s = \frac{5}{2} rs = -9$$

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

$$r = \frac{5}{4} - u s = \frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -9

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -9$$

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

$$\frac{25}{16} - u^2 = -9$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -9-\frac{25}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =\frac{5}{4} - \frac{13}{4} = -2 s = \frac{5}{4} + \frac{13}{4} = 4.500$$

Evaluate

$$6\left(2x-9\right)\left(x+2\right)$$