Solve 12n^2-50n+50

$$12 n ^ { 2 } - 50 n + 50$$

Factor

$$2\left(2n-5\right)\left(3n-5\right)$$

Solution Steps

Factor out 2.

$$2\left(6n^{2}-25n+25\right)$$

Consider 6n^{2}-25n+25. Factor the expression by grouping. First, the expression needs to be rewritten as 6n^{2}+an+bn+25. To find a and b, set up a system to be solved.

$$a+b=-25 ab=6\times 25=150$$

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 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -25.

$$a=-15 b=-10$$

Rewrite 6n^{2}-25n+25 as \left(6n^{2}-15n\right)+\left(-10n+25\right).

$$\left(6n^{2}-15n\right)+\left(-10n+25\right)$$

Factor out 3n in the first and -5 in the second group.

$$3n\left(2n-5\right)-5\left(2n-5\right)$$

Factor out common term 2n-5 by using distributive property.

$$\left(2n-5\right)\left(3n-5\right)$$

Rewrite the complete factored expression.

$$2\left(2n-5\right)\left(3n-5\right)$$

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.

$$12n^{2}-50n+50=0$$

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.

$$n=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 12\times 50}}{2\times 12}$$

Square -50.

$$n=\frac{-\left(-50\right)±\sqrt{2500-4\times 12\times 50}}{2\times 12}$$

Multiply -4 times 12.

$$n=\frac{-\left(-50\right)±\sqrt{2500-48\times 50}}{2\times 12}$$

Multiply -48 times 50.

$$n=\frac{-\left(-50\right)±\sqrt{2500-2400}}{2\times 12}$$

Add 2500 to -2400.

$$n=\frac{-\left(-50\right)±\sqrt{100}}{2\times 12}$$

Take the square root of 100.

$$n=\frac{-\left(-50\right)±10}{2\times 12}$$

The opposite of -50 is 50.

$$n=\frac{50±10}{2\times 12}$$

Multiply 2 times 12.

$$n=\frac{50±10}{24}$$

Now solve the equation n=\frac{50±10}{24} when ± is plus. Add 50 to 10.

$$n=\frac{60}{24}$$

Reduce the fraction \frac{60}{24} to lowest terms by extracting and canceling out 12.

$$n=\frac{5}{2}$$

Now solve the equation n=\frac{50±10}{24} when ± is minus. Subtract 10 from 50.

$$n=\frac{40}{24}$$

Reduce the fraction \frac{40}{24} to lowest terms by extracting and canceling out 8.

$$n=\frac{5}{3}$$

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

$$12n^{2}-50n+50=12\left(n-\frac{5}{2}\right)\left(n-\frac{5}{3}\right)$$

Subtract \frac{5}{2} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12n^{2}-50n+50=12\times \frac{2n-5}{2}\left(n-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12n^{2}-50n+50=12\times \frac{2n-5}{2}\times \frac{3n-5}{3}$$

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

$$12n^{2}-50n+50=12\times \frac{\left(2n-5\right)\left(3n-5\right)}{2\times 3}$$

Multiply 2 times 3.

$$12n^{2}-50n+50=12\times \frac{\left(2n-5\right)\left(3n-5\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12n^{2}-50n+50=2\left(2n-5\right)\left(3n-5\right)$$

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 12

$$x ^ 2 -\frac{25}{6}x +\frac{25}{6} = 0$$

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

$$r + s = \frac{25}{6} rs = \frac{25}{6}$$

Two numbers r and s sum up to \frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{6} = \frac{25}{12}. 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>

$$r = \frac{25}{12} - u s = \frac{25}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{6}

$$(\frac{25}{12} - u) (\frac{25}{12} + u) = \frac{25}{6}$$

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

$$\frac{625}{144} - u^2 = \frac{25}{6}$$

Simplify the expression by subtracting \frac{625}{144} on both sides

$$-u^2 = \frac{25}{6}-\frac{625}{144} = -\frac{25}{144}$$

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

$$u^2 = \frac{25}{144} u = \pm\sqrt{\frac{25}{144}} = \pm \frac{5}{12}$$

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

$$r =\frac{25}{12} - \frac{5}{12} = 1.667 s = \frac{25}{12} + \frac{5}{12} = 2.500$$

Evaluate

$$12n^{2}-50n+50$$