Solve 12x^2-20x-48

$$12 x ^ { 2 } - 20 x - 48$$

Factor

$$4\left(x-3\right)\left(3x+4\right)$$

Solution Steps

Factor out 4.

$$4\left(3x^{2}-5x-12\right)$$

Consider 3x^{2}-5x-12. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-5 ab=3\left(-12\right)=-36$$

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 3x^{2}-5x-12 as \left(3x^{2}-9x\right)+\left(4x-12\right).

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

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-3\right)+4\left(x-3\right)$$

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

$$\left(x-3\right)\left(3x+4\right)$$

Rewrite the complete factored expression.

$$4\left(x-3\right)\left(3x+4\right)$$

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.

$$12x^{2}-20x-48=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 12\left(-48\right)}}{2\times 12}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 12\left(-48\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-20\right)±\sqrt{400-48\left(-48\right)}}{2\times 12}$$

Multiply -48 times -48.

$$x=\frac{-\left(-20\right)±\sqrt{400+2304}}{2\times 12}$$

Add 400 to 2304.

$$x=\frac{-\left(-20\right)±\sqrt{2704}}{2\times 12}$$

Take the square root of 2704.

$$x=\frac{-\left(-20\right)±52}{2\times 12}$$

The opposite of -20 is 20.

$$x=\frac{20±52}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{20±52}{24}$$

Now solve the equation x=\frac{20±52}{24} when ± is plus. Add 20 to 52.

$$x=\frac{72}{24}$$

Divide 72 by 24.

$$x=3$$

Now solve the equation x=\frac{20±52}{24} when ± is minus. Subtract 52 from 20.

$$x=-\frac{32}{24}$$

Reduce the fraction \frac{-32}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -\frac{4}{3} for x_{2}.

$$12x^{2}-20x-48=12\left(x-3\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$12x^{2}-20x-48=12\left(x-3\right)\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-20x-48=12\left(x-3\right)\times \frac{3x+4}{3}$$

Cancel out 3, the greatest common factor in 12 and 3.

$$12x^{2}-20x-48=4\left(x-3\right)\left(3x+4\right)$$

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

$$x ^ 2 -\frac{5}{3}x -4 = 0$$

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

$$r + s = \frac{5}{3} rs = -4$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = -4$$

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

$$\frac{25}{36} - u^2 = -4$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = -4-\frac{25}{36} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =\frac{5}{6} - \frac{13}{6} = -1.333 s = \frac{5}{6} + \frac{13}{6} = 3$$

Evaluate

$$4\left(x-3\right)\left(3x+4\right)$$