Solve 5n^2-75n+280

$$5 n ^ { 2 } - 75 n + 280$$

Factor

$$5\left(n-8\right)\left(n-7\right)$$

Solution Steps

Factor out 5.

$$5\left(n^{2}-15n+56\right)$$

Consider n^{2}-15n+56. Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+56. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 56=56$$

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 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -15.

$$a=-8 b=-7$$

Rewrite n^{2}-15n+56 as \left(n^{2}-8n\right)+\left(-7n+56\right).

$$\left(n^{2}-8n\right)+\left(-7n+56\right)$$

Factor out n in the first and -7 in the second group.

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

Factor out common term n-8 by using distributive property.

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

Rewrite the complete factored expression.

$$5\left(n-8\right)\left(n-7\right)$$

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.

$$5n^{2}-75n+280=0$$

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.

$$n=\frac{-\left(-75\right)±\sqrt{\left(-75\right)^{2}-4\times 5\times 280}}{2\times 5}$$

Square -75.

$$n=\frac{-\left(-75\right)±\sqrt{5625-4\times 5\times 280}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-\left(-75\right)±\sqrt{5625-20\times 280}}{2\times 5}$$

Multiply -20 times 280.

$$n=\frac{-\left(-75\right)±\sqrt{5625-5600}}{2\times 5}$$

Add 5625 to -5600.

$$n=\frac{-\left(-75\right)±\sqrt{25}}{2\times 5}$$

Take the square root of 25.

$$n=\frac{-\left(-75\right)±5}{2\times 5}$$

The opposite of -75 is 75.

$$n=\frac{75±5}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{75±5}{10}$$

Now solve the equation n=\frac{75±5}{10} when ± is plus. Add 75 to 5.

$$n=\frac{80}{10}$$

Divide 80 by 10.

$$n=8$$

Now solve the equation n=\frac{75±5}{10} when ± is minus. Subtract 5 from 75.

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

Divide 70 by 10.

$$n=7$$

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

$$5n^{2}-75n+280=5\left(n-8\right)\left(n-7\right)$$

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

$$x ^ 2 -15x +56 = 0$$

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

$$r + s = 15 rs = 56$$

Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{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>

$$r = \frac{15}{2} - u s = \frac{15}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 56

$$(\frac{15}{2} - u) (\frac{15}{2} + u) = 56$$

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

$$\frac{225}{4} - u^2 = 56$$

Simplify the expression by subtracting \frac{225}{4} on both sides

$$-u^2 = 56-\frac{225}{4} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{15}{2} - \frac{1}{2} = 7 s = \frac{15}{2} + \frac{1}{2} = 8$$

Evaluate

$$5\left(n-8\right)\left(n-7\right)$$