Solve 9n^2-64n+60

$$9 n ^ { 2 } - 64 n + 60$$

Factor

$$\left(n-6\right)\left(9n-10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9n^{2}+an+bn+60. To find a and b, set up a system to be solved.

$$a+b=-64 ab=9\times 60=540$$

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 540.

$$-1,-540 -2,-270 -3,-180 -4,-135 -5,-108 -6,-90 -9,-60 -10,-54 -12,-45 -15,-36 -18,-30 -20,-27$$

Calculate the sum for each pair.

$$-1-540=-541 -2-270=-272 -3-180=-183 -4-135=-139 -5-108=-113 -6-90=-96 -9-60=-69 -10-54=-64 -12-45=-57 -15-36=-51 -18-30=-48 -20-27=-47$$

The solution is the pair that gives sum -64.

$$a=-54 b=-10$$

Rewrite 9n^{2}-64n+60 as \left(9n^{2}-54n\right)+\left(-10n+60\right).

$$\left(9n^{2}-54n\right)+\left(-10n+60\right)$$

Factor out 9n in the first and -10 in the second group.

$$9n\left(n-6\right)-10\left(n-6\right)$$

Factor out common term n-6 by using distributive property.

$$\left(n-6\right)\left(9n-10\right)$$

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.

$$9n^{2}-64n+60=0$$

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.

$$n=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\times 9\times 60}}{2\times 9}$$

Square -64.

$$n=\frac{-\left(-64\right)±\sqrt{4096-4\times 9\times 60}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{-\left(-64\right)±\sqrt{4096-36\times 60}}{2\times 9}$$

Multiply -36 times 60.

$$n=\frac{-\left(-64\right)±\sqrt{4096-2160}}{2\times 9}$$

Add 4096 to -2160.

$$n=\frac{-\left(-64\right)±\sqrt{1936}}{2\times 9}$$

Take the square root of 1936.

$$n=\frac{-\left(-64\right)±44}{2\times 9}$$

The opposite of -64 is 64.

$$n=\frac{64±44}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{64±44}{18}$$

Now solve the equation n=\frac{64±44}{18} when ± is plus. Add 64 to 44.

$$n=\frac{108}{18}$$

Divide 108 by 18.

$$n=6$$

Now solve the equation n=\frac{64±44}{18} when ± is minus. Subtract 44 from 64.

$$n=\frac{20}{18}$$

Reduce the fraction \frac{20}{18} to lowest terms by extracting and canceling out 2.

$$n=\frac{10}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and \frac{10}{9} for x_{2}.

$$9n^{2}-64n+60=9\left(n-6\right)\left(n-\frac{10}{9}\right)$$

Subtract \frac{10}{9} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9n^{2}-64n+60=9\left(n-6\right)\times \frac{9n-10}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9n^{2}-64n+60=\left(n-6\right)\left(9n-10\right)$$

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 9

$$x ^ 2 -\frac{64}{9}x +\frac{20}{3} = 0$$

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

$$r + s = \frac{64}{9} rs = \frac{20}{3}$$

Two numbers r and s sum up to \frac{64}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{64}{9} = \frac{32}{9}. 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>

$$r = \frac{32}{9} - u s = \frac{32}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{3}

$$(\frac{32}{9} - u) (\frac{32}{9} + u) = \frac{20}{3}$$

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

$$\frac{1024}{81} - u^2 = \frac{20}{3}$$

Simplify the expression by subtracting \frac{1024}{81} on both sides

$$-u^2 = \frac{20}{3}-\frac{1024}{81} = -\frac{484}{81}$$

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

$$u^2 = \frac{484}{81} u = \pm\sqrt{\frac{484}{81}} = \pm \frac{22}{9}$$

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

$$r =\frac{32}{9} - \frac{22}{9} = 1.111 s = \frac{32}{9} + \frac{22}{9} = 6$$

Evaluate

$$\left(n-6\right)\left(9n-10\right)$$