Solve 24x^2+x-10

$$24 x ^ { 2 } + x - 10$$

Factor

$$\left(8x-5\right)\left(3x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=1 ab=24\left(-10\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 1.

$$a=-15 b=16$$

Rewrite 24x^{2}+x-10 as \left(24x^{2}-15x\right)+\left(16x-10\right).

$$\left(24x^{2}-15x\right)+\left(16x-10\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(8x-5\right)+2\left(8x-5\right)$$

Factor out common term 8x-5 by using distributive property.

$$\left(8x-5\right)\left(3x+2\right)$$

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.

$$24x^{2}+x-10=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 24\left(-10\right)}}{2\times 24}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 24\left(-10\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-1±\sqrt{1-96\left(-10\right)}}{2\times 24}$$

Multiply -96 times -10.

$$x=\frac{-1±\sqrt{1+960}}{2\times 24}$$

Add 1 to 960.

$$x=\frac{-1±\sqrt{961}}{2\times 24}$$

Take the square root of 961.

$$x=\frac{-1±31}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-1±31}{48}$$

Now solve the equation x=\frac{-1±31}{48} when ± is plus. Add -1 to 31.

$$x=\frac{30}{48}$$

Reduce the fraction \frac{30}{48} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{8}$$

Now solve the equation x=\frac{-1±31}{48} when ± is minus. Subtract 31 from -1.

$$x=-\frac{32}{48}$$

Reduce the fraction \frac{-32}{48} to lowest terms by extracting and canceling out 16.

$$x=-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{8} for x_{1} and -\frac{2}{3} for x_{2}.

$$24x^{2}+x-10=24\left(x-\frac{5}{8}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$24x^{2}+x-10=24\left(x-\frac{5}{8}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{5}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+x-10=24\times \frac{8x-5}{8}\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+x-10=24\times \frac{8x-5}{8}\times \frac{3x+2}{3}$$

Multiply \frac{8x-5}{8} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24x^{2}+x-10=24\times \frac{\left(8x-5\right)\left(3x+2\right)}{8\times 3}$$

Multiply 8 times 3.

$$24x^{2}+x-10=24\times \frac{\left(8x-5\right)\left(3x+2\right)}{24}$$

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

$$24x^{2}+x-10=\left(8x-5\right)\left(3x+2\right)$$

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

$$x ^ 2 +\frac{1}{24}x -\frac{5}{12} = 0$$

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

$$r + s = -\frac{1}{24} rs = -\frac{5}{12}$$

Two numbers r and s sum up to -\frac{1}{24} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{24} = -\frac{1}{48}. 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>

$$r = -\frac{1}{48} - u s = -\frac{1}{48} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{12}

$$(-\frac{1}{48} - u) (-\frac{1}{48} + u) = -\frac{5}{12}$$

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

$$\frac{1}{2304} - u^2 = -\frac{5}{12}$$

Simplify the expression by subtracting \frac{1}{2304} on both sides

$$-u^2 = -\frac{5}{12}-\frac{1}{2304} = -\frac{961}{2304}$$

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

$$u^2 = \frac{961}{2304} u = \pm\sqrt{\frac{961}{2304}} = \pm \frac{31}{48}$$

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

$$r =-\frac{1}{48} - \frac{31}{48} = -0.667 s = -\frac{1}{48} + \frac{31}{48} = 0.625$$

Evaluate

$$24x^{2}+x-10$$