Solve 12x^2+23x-24

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

Factor

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-9 b=32$$

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

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

Factor out 3x in the first and 8 in the second group.

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

Factor out common term 4x-3 by using distributive property.

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

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.

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

Square 23.

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

Multiply -4 times 12.

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

Multiply -48 times -24.

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

Add 529 to 1152.

$$x=\frac{-23±\sqrt{1681}}{2\times 12}$$

Take the square root of 1681.

$$x=\frac{-23±41}{2\times 12}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply 4 times 3.

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

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

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

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.

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

Evaluate

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