Solve 24n^2-65n+21

$$24 n ^ { 2 } - 65 n + 21$$

Factor

$$\left(3n-7\right)\left(8n-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 24n^{2}+an+bn+21. To find a and b, set up a system to be solved.

$$a+b=-65 ab=24\times 21=504$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 504.

$$-1,-504 -2,-252 -3,-168 -4,-126 -6,-84 -7,-72 -8,-63 -9,-56 -12,-42 -14,-36 -18,-28 -21,-24$$

Calculate the sum for each pair.

$$-1-504=-505 -2-252=-254 -3-168=-171 -4-126=-130 -6-84=-90 -7-72=-79 -8-63=-71 -9-56=-65 -12-42=-54 -14-36=-50 -18-28=-46 -21-24=-45$$

The solution is the pair that gives sum -65.

$$a=-56 b=-9$$

Rewrite 24n^{2}-65n+21 as \left(24n^{2}-56n\right)+\left(-9n+21\right).

$$\left(24n^{2}-56n\right)+\left(-9n+21\right)$$

Factor out 8n in the first and -3 in the second group.

$$8n\left(3n-7\right)-3\left(3n-7\right)$$

Factor out common term 3n-7 by using distributive property.

$$\left(3n-7\right)\left(8n-3\right)$$

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.

$$24n^{2}-65n+21=0$$

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.

$$n=\frac{-\left(-65\right)±\sqrt{\left(-65\right)^{2}-4\times 24\times 21}}{2\times 24}$$

Square -65.

$$n=\frac{-\left(-65\right)±\sqrt{4225-4\times 24\times 21}}{2\times 24}$$

Multiply -4 times 24.

$$n=\frac{-\left(-65\right)±\sqrt{4225-96\times 21}}{2\times 24}$$

Multiply -96 times 21.

$$n=\frac{-\left(-65\right)±\sqrt{4225-2016}}{2\times 24}$$

Add 4225 to -2016.

$$n=\frac{-\left(-65\right)±\sqrt{2209}}{2\times 24}$$

Take the square root of 2209.

$$n=\frac{-\left(-65\right)±47}{2\times 24}$$

The opposite of -65 is 65.

$$n=\frac{65±47}{2\times 24}$$

Multiply 2 times 24.

$$n=\frac{65±47}{48}$$

Now solve the equation n=\frac{65±47}{48} when ± is plus. Add 65 to 47.

$$n=\frac{112}{48}$$

Reduce the fraction \frac{112}{48} to lowest terms by extracting and canceling out 16.

$$n=\frac{7}{3}$$

Now solve the equation n=\frac{65±47}{48} when ± is minus. Subtract 47 from 65.

$$n=\frac{18}{48}$$

Reduce the fraction \frac{18}{48} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{3} for x_{1} and \frac{3}{8} for x_{2}.

$$24n^{2}-65n+21=24\left(n-\frac{7}{3}\right)\left(n-\frac{3}{8}\right)$$

Subtract \frac{7}{3} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$24n^{2}-65n+21=24\times \frac{3n-7}{3}\left(n-\frac{3}{8}\right)$$

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

$$24n^{2}-65n+21=24\times \frac{3n-7}{3}\times \frac{8n-3}{8}$$

Multiply \frac{3n-7}{3} times \frac{8n-3}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$24n^{2}-65n+21=24\times \frac{\left(3n-7\right)\left(8n-3\right)}{3\times 8}$$

Multiply 3 times 8.

$$24n^{2}-65n+21=24\times \frac{\left(3n-7\right)\left(8n-3\right)}{24}$$

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

$$24n^{2}-65n+21=\left(3n-7\right)\left(8n-3\right)$$

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 24

$$x ^ 2 -\frac{65}{24}x +\frac{7}{8} = 0$$

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

$$r + s = \frac{65}{24} rs = \frac{7}{8}$$

Two numbers r and s sum up to \frac{65}{24} exactly when the average of the two numbers is \frac{1}{2}*\frac{65}{24} = \frac{65}{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>

$$r = \frac{65}{48} - u s = \frac{65}{48} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{8}

$$(\frac{65}{48} - u) (\frac{65}{48} + u) = \frac{7}{8}$$

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

$$\frac{4225}{2304} - u^2 = \frac{7}{8}$$

Simplify the expression by subtracting \frac{4225}{2304} on both sides

$$-u^2 = \frac{7}{8}-\frac{4225}{2304} = -\frac{2209}{2304}$$

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

$$u^2 = \frac{2209}{2304} u = \pm\sqrt{\frac{2209}{2304}} = \pm \frac{47}{48}$$

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

$$r =\frac{65}{48} - \frac{47}{48} = 0.375 s = \frac{65}{48} + \frac{47}{48} = 2.333$$

Evaluate

$$\left(3n-7\right)\left(8n-3\right)$$