Solve 12x^2-23x-24

$$12 x ^ { 2 } - 23 x - 24$$

Factor

$$\left(3x-8\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=-23 ab=12\left(-24\right)=-288$$

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 -288.

$$1,-288 2,-144 3,-96 4,-72 6,-48 8,-36 9,-32 12,-24 16,-18$$

Calculate the sum for each pair.

$$1-288=-287 2-144=-142 3-96=-93 4-72=-68 6-48=-42 8-36=-28 9-32=-23 12-24=-12 16-18=-2$$

The solution is the pair that gives sum -23.

$$a=-32 b=9$$

Rewrite 12x^{2}-23x-24 as \left(12x^{2}-32x\right)+\left(9x-24\right).

$$\left(12x^{2}-32x\right)+\left(9x-24\right)$$

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

$$4x\left(3x-8\right)+3\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(4x+3\right)$$

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.

$$12x^{2}-23x-24=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 12\left(-24\right)}}{2\times 12}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 12\left(-24\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-23\right)±\sqrt{529-48\left(-24\right)}}{2\times 12}$$

Multiply -48 times -24.

$$x=\frac{-\left(-23\right)±\sqrt{529+1152}}{2\times 12}$$

Add 529 to 1152.

$$x=\frac{-\left(-23\right)±\sqrt{1681}}{2\times 12}$$

Take the square root of 1681.

$$x=\frac{-\left(-23\right)±41}{2\times 12}$$

The opposite of -23 is 23.

$$x=\frac{23±41}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{23±41}{24}$$

Now solve the equation x=\frac{23±41}{24} when ± is plus. Add 23 to 41.

$$x=\frac{64}{24}$$

Reduce the fraction \frac{64}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{23±41}{24} when ± is minus. Subtract 41 from 23.

$$x=-\frac{18}{24}$$

Reduce the fraction \frac{-18}{24} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and -\frac{3}{4} for x_{2}.

$$12x^{2}-23x-24=12\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$12x^{2}-23x-24=12\left(x-\frac{8}{3}\right)\left(x+\frac{3}{4}\right)$$

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-23x-24=12\times \frac{3x-8}{3}\left(x+\frac{3}{4}\right)$$

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

$$12x^{2}-23x-24=12\times \frac{3x-8}{3}\times \frac{4x+3}{4}$$

Multiply \frac{3x-8}{3} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-23x-24=12\times \frac{\left(3x-8\right)\left(4x+3\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}-23x-24=12\times \frac{\left(3x-8\right)\left(4x+3\right)}{12}$$

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

$$12x^{2}-23x-24=\left(3x-8\right)\left(4x+3\right)$$

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 12

$$x ^ 2 -\frac{23}{12}x -2 = 0$$

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

$$r + s = \frac{23}{12} rs = -2$$

Two numbers r and s sum up to \frac{23}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{12} = \frac{23}{24}. 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>

$$r = \frac{23}{24} - u s = \frac{23}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -2

$$(\frac{23}{24} - u) (\frac{23}{24} + u) = -2$$

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

$$\frac{529}{576} - u^2 = -2$$

Simplify the expression by subtracting \frac{529}{576} on both sides

$$-u^2 = -2-\frac{529}{576} = -\frac{1681}{576}$$

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

$$u^2 = \frac{1681}{576} u = \pm\sqrt{\frac{1681}{576}} = \pm \frac{41}{24}$$

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

$$r =\frac{23}{24} - \frac{41}{24} = -0.750 s = \frac{23}{24} + \frac{41}{24} = 2.667$$

Evaluate

$$\left(3x-8\right)\left(4x+3\right)$$