Solve 24x^2-26x-8

$$24 x ^ { 2 } - 26 x - 8$$

Factor

$$2\left(3x-4\right)\left(4x+1\right)$$

Solution Steps

Factor out 2.

$$2\left(12x^{2}-13x-4\right)$$

Consider 12x^{2}-13x-4. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-13 ab=12\left(-4\right)=-48$$

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 -48.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -13.

$$a=-16 b=3$$

Rewrite 12x^{2}-13x-4 as \left(12x^{2}-16x\right)+\left(3x-4\right).

$$\left(12x^{2}-16x\right)+\left(3x-4\right)$$

Factor out 4x in 12x^{2}-16x.

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

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

$$\left(3x-4\right)\left(4x+1\right)$$

Rewrite the complete factored expression.

$$2\left(3x-4\right)\left(4x+1\right)$$

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.

$$24x^{2}-26x-8=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 24\left(-8\right)}}{2\times 24}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 24\left(-8\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-26\right)±\sqrt{676-96\left(-8\right)}}{2\times 24}$$

Multiply -96 times -8.

$$x=\frac{-\left(-26\right)±\sqrt{676+768}}{2\times 24}$$

Add 676 to 768.

$$x=\frac{-\left(-26\right)±\sqrt{1444}}{2\times 24}$$

Take the square root of 1444.

$$x=\frac{-\left(-26\right)±38}{2\times 24}$$

The opposite of -26 is 26.

$$x=\frac{26±38}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{26±38}{48}$$

Now solve the equation x=\frac{26±38}{48} when ± is plus. Add 26 to 38.

$$x=\frac{64}{48}$$

Reduce the fraction \frac{64}{48} to lowest terms by extracting and canceling out 16.

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

Now solve the equation x=\frac{26±38}{48} when ± is minus. Subtract 38 from 26.

$$x=-\frac{12}{48}$$

Reduce the fraction \frac{-12}{48} to lowest terms by extracting and canceling out 12.

$$x=-\frac{1}{4}$$

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

$$24x^{2}-26x-8=24\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

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

$$24x^{2}-26x-8=24\left(x-\frac{4}{3}\right)\left(x+\frac{1}{4}\right)$$

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

$$24x^{2}-26x-8=24\times \frac{3x-4}{3}\left(x+\frac{1}{4}\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}-26x-8=24\times \frac{3x-4}{3}\times \frac{4x+1}{4}$$

Multiply \frac{3x-4}{3} times \frac{4x+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}-26x-8=24\times \frac{\left(3x-4\right)\left(4x+1\right)}{3\times 4}$$

Multiply 3 times 4.

$$24x^{2}-26x-8=24\times \frac{\left(3x-4\right)\left(4x+1\right)}{12}$$

Cancel out 12, the greatest common factor in 24 and 12.

$$24x^{2}-26x-8=2\left(3x-4\right)\left(4x+1\right)$$

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

$$x ^ 2 -\frac{13}{12}x -\frac{1}{3} = 0$$

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

$$r + s = \frac{13}{12} rs = -\frac{1}{3}$$

Two numbers r and s sum up to \frac{13}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{12} = \frac{13}{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>

$$r = \frac{13}{24} - u s = \frac{13}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(\frac{13}{24} - u) (\frac{13}{24} + u) = -\frac{1}{3}$$

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

$$\frac{169}{576} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{169}{576} on both sides

$$-u^2 = -\frac{1}{3}-\frac{169}{576} = -\frac{361}{576}$$

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

$$u^2 = \frac{361}{576} u = \pm\sqrt{\frac{361}{576}} = \pm \frac{19}{24}$$

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

$$r =\frac{13}{24} - \frac{19}{24} = -0.250 s = \frac{13}{24} + \frac{19}{24} = 1.333$$

Evaluate

$$24x^{2}-26x-8$$