Solve 25n^2+160n+60

$$25 n ^ { 2 } + 160 n + 60$$

Factor

$$5\left(n+6\right)\left(5n+2\right)$$

Solution Steps

Factor out 5.

$$5\left(5n^{2}+32n+12\right)$$

Consider 5n^{2}+32n+12. Factor the expression by grouping. First, the expression needs to be rewritten as 5n^{2}+an+bn+12. To find a and b, set up a system to be solved.

$$a+b=32 ab=5\times 12=60$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 32.

$$a=2 b=30$$

Rewrite 5n^{2}+32n+12 as \left(5n^{2}+2n\right)+\left(30n+12\right).

$$\left(5n^{2}+2n\right)+\left(30n+12\right)$$

Factor out n in the first and 6 in the second group.

$$n\left(5n+2\right)+6\left(5n+2\right)$$

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

$$\left(5n+2\right)\left(n+6\right)$$

Rewrite the complete factored expression.

$$5\left(5n+2\right)\left(n+6\right)$$

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.

$$25n^{2}+160n+60=0$$

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.

$$n=\frac{-160±\sqrt{160^{2}-4\times 25\times 60}}{2\times 25}$$

Square 160.

$$n=\frac{-160±\sqrt{25600-4\times 25\times 60}}{2\times 25}$$

Multiply -4 times 25.

$$n=\frac{-160±\sqrt{25600-100\times 60}}{2\times 25}$$

Multiply -100 times 60.

$$n=\frac{-160±\sqrt{25600-6000}}{2\times 25}$$

Add 25600 to -6000.

$$n=\frac{-160±\sqrt{19600}}{2\times 25}$$

Take the square root of 19600.

$$n=\frac{-160±140}{2\times 25}$$

Multiply 2 times 25.

$$n=\frac{-160±140}{50}$$

Now solve the equation n=\frac{-160±140}{50} when ± is plus. Add -160 to 140.

$$n=-\frac{20}{50}$$

Reduce the fraction \frac{-20}{50} to lowest terms by extracting and canceling out 10.

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

Now solve the equation n=\frac{-160±140}{50} when ± is minus. Subtract 140 from -160.

$$n=-\frac{300}{50}$$

Divide -300 by 50.

$$n=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{5} for x_{1} and -6 for x_{2}.

$$25n^{2}+160n+60=25\left(n-\left(-\frac{2}{5}\right)\right)\left(n-\left(-6\right)\right)$$

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

$$25n^{2}+160n+60=25\left(n+\frac{2}{5}\right)\left(n+6\right)$$

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

$$25n^{2}+160n+60=25\times \frac{5n+2}{5}\left(n+6\right)$$

Cancel out 5, the greatest common factor in 25 and 5.

$$25n^{2}+160n+60=5\left(5n+2\right)\left(n+6\right)$$

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 25

$$x ^ 2 +\frac{32}{5}x +\frac{12}{5} = 0$$

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

$$r + s = -\frac{32}{5} rs = \frac{12}{5}$$

Two numbers r and s sum up to -\frac{32}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{32}{5} = -\frac{16}{5}. 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>

$$r = -\frac{16}{5} - u s = -\frac{16}{5} + u$$

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

$$(-\frac{16}{5} - u) (-\frac{16}{5} + u) = \frac{12}{5}$$

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

$$\frac{256}{25} - u^2 = \frac{12}{5}$$

Simplify the expression by subtracting \frac{256}{25} on both sides

$$-u^2 = \frac{12}{5}-\frac{256}{25} = -\frac{196}{25}$$

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

$$u^2 = \frac{196}{25} u = \pm\sqrt{\frac{196}{25}} = \pm \frac{14}{5}$$

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

$$r =-\frac{16}{5} - \frac{14}{5} = -6 s = -\frac{16}{5} + \frac{14}{5} = -0.400$$

Evaluate

$$5\left(n+6\right)\left(5n+2\right)$$