Solve 5n^2-55n+140=

$$5 n ^ { 2 } - 55 n + 140 =$$

Factor

$$5\left(n-7\right)\left(n-4\right)$$

Solution Steps

Factor out 5.

$$5\left(n^{2}-11n+28\right)$$

Consider n^{2}-11n+28. Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+28. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 28=28$$

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 28.

$$-1,-28 -2,-14 -4,-7$$

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -11.

$$a=-7 b=-4$$

Rewrite n^{2}-11n+28 as \left(n^{2}-7n\right)+\left(-4n+28\right).

$$\left(n^{2}-7n\right)+\left(-4n+28\right)$$

Factor out n in the first and -4 in the second group.

$$n\left(n-7\right)-4\left(n-7\right)$$

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

$$\left(n-7\right)\left(n-4\right)$$

Rewrite the complete factored expression.

$$5\left(n-7\right)\left(n-4\right)$$

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.

$$5n^{2}-55n+140=0$$

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.

$$n=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\times 5\times 140}}{2\times 5}$$

Square -55.

$$n=\frac{-\left(-55\right)±\sqrt{3025-4\times 5\times 140}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-\left(-55\right)±\sqrt{3025-20\times 140}}{2\times 5}$$

Multiply -20 times 140.

$$n=\frac{-\left(-55\right)±\sqrt{3025-2800}}{2\times 5}$$

Add 3025 to -2800.

$$n=\frac{-\left(-55\right)±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$n=\frac{-\left(-55\right)±15}{2\times 5}$$

The opposite of -55 is 55.

$$n=\frac{55±15}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{55±15}{10}$$

Now solve the equation n=\frac{55±15}{10} when ± is plus. Add 55 to 15.

$$n=\frac{70}{10}$$

Divide 70 by 10.

$$n=7$$

Now solve the equation n=\frac{55±15}{10} when ± is minus. Subtract 15 from 55.

$$n=\frac{40}{10}$$

Divide 40 by 10.

$$n=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and 4 for x_{2}.

$$5n^{2}-55n+140=5\left(n-7\right)\left(n-4\right)$$

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 5

$$x ^ 2 -11x +28 = 0$$

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

$$r + s = 11 rs = 28$$

Two numbers r and s sum up to 11 exactly when the average of the two numbers is \frac{1}{2}*11 = \frac{11}{2}. 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>

$$r = \frac{11}{2} - u s = \frac{11}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 28

$$(\frac{11}{2} - u) (\frac{11}{2} + u) = 28$$

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

$$\frac{121}{4} - u^2 = 28$$

Simplify the expression by subtracting \frac{121}{4} on both sides

$$-u^2 = 28-\frac{121}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =\frac{11}{2} - \frac{3}{2} = 4 s = \frac{11}{2} + \frac{3}{2} = 7$$

Evaluate

$$5\left(n-7\right)\left(n-4\right)$$