Solve 24x^2-60x-36

$$24 x ^ { 2 } - 60 x - 36$$

Factor

$$12\left(x-3\right)\left(2x+1\right)$$

Solution Steps

Factor out 12.

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

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

$$a+b=-5 ab=2\left(-3\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 2x^{2}-5x-3 as \left(2x^{2}-6x\right)+\left(x-3\right).

$$\left(2x^{2}-6x\right)+\left(x-3\right)$$

Factor out 2x in 2x^{2}-6x.

$$2x\left(x-3\right)+x-3$$

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

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

Rewrite the complete factored expression.

$$12\left(x-3\right)\left(2x+1\right)$$

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.

$$24x^{2}-60x-36=0$$

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.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 24\left(-36\right)}}{2\times 24}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 24\left(-36\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-60\right)±\sqrt{3600-96\left(-36\right)}}{2\times 24}$$

Multiply -96 times -36.

$$x=\frac{-\left(-60\right)±\sqrt{3600+3456}}{2\times 24}$$

Add 3600 to 3456.

$$x=\frac{-\left(-60\right)±\sqrt{7056}}{2\times 24}$$

Take the square root of 7056.

$$x=\frac{-\left(-60\right)±84}{2\times 24}$$

The opposite of -60 is 60.

$$x=\frac{60±84}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{60±84}{48}$$

Now solve the equation x=\frac{60±84}{48} when ± is plus. Add 60 to 84.

$$x=\frac{144}{48}$$

Divide 144 by 48.

$$x=3$$

Now solve the equation x=\frac{60±84}{48} when ± is minus. Subtract 84 from 60.

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

Reduce the fraction \frac{-24}{48} to lowest terms by extracting and canceling out 24.

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

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

$$24x^{2}-60x-36=24\left(x-3\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$24x^{2}-60x-36=24\left(x-3\right)\left(x+\frac{1}{2}\right)$$

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

$$24x^{2}-60x-36=24\left(x-3\right)\times \frac{2x+1}{2}$$

Cancel out 2, the greatest common factor in 24 and 2.

$$24x^{2}-60x-36=12\left(x-3\right)\left(2x+1\right)$$

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 24

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

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

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

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 = -\frac{3}{2}

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = -\frac{3}{2}$$

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

$$\frac{25}{16} - u^2 = -\frac{3}{2}$$

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

$$-u^2 = -\frac{3}{2}-\frac{25}{16} = -\frac{49}{16}$$

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

$$u^2 = \frac{49}{16} u = \pm\sqrt{\frac{49}{16}} = \pm \frac{7}{4}$$

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

$$r =\frac{5}{4} - \frac{7}{4} = -0.500 s = \frac{5}{4} + \frac{7}{4} = 3$$

Evaluate

$$12\left(x-3\right)\left(2x+1\right)$$