Solve 3n^2-n-520

$$3 n ^ { 2 } - n - 520$$

Factor

$$\left(3n-40\right)\left(n+13\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3n^{2}+an+bn-520. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-520\right)=-1560$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1560.

$$1,-1560 2,-780 3,-520 4,-390 5,-312 6,-260 8,-195 10,-156 12,-130 13,-120 15,-104 20,-78 24,-65 26,-60 30,-52 39,-40$$

Calculate the sum for each pair.

$$1-1560=-1559 2-780=-778 3-520=-517 4-390=-386 5-312=-307 6-260=-254 8-195=-187 10-156=-146 12-130=-118 13-120=-107 15-104=-89 20-78=-58 24-65=-41 26-60=-34 30-52=-22 39-40=-1$$

The solution is the pair that gives sum -1.

$$a=-40 b=39$$

Rewrite 3n^{2}-n-520 as \left(3n^{2}-40n\right)+\left(39n-520\right).

$$\left(3n^{2}-40n\right)+\left(39n-520\right)$$

Factor out n in the first and 13 in the second group.

$$n\left(3n-40\right)+13\left(3n-40\right)$$

Factor out common term 3n-40 by using distributive property.

$$\left(3n-40\right)\left(n+13\right)$$

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.

$$3n^{2}-n-520=0$$

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.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-520\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-1\right)±\sqrt{1-12\left(-520\right)}}{2\times 3}$$

Multiply -12 times -520.

$$n=\frac{-\left(-1\right)±\sqrt{1+6240}}{2\times 3}$$

Add 1 to 6240.

$$n=\frac{-\left(-1\right)±\sqrt{6241}}{2\times 3}$$

Take the square root of 6241.

$$n=\frac{-\left(-1\right)±79}{2\times 3}$$

The opposite of -1 is 1.

$$n=\frac{1±79}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{1±79}{6}$$

Now solve the equation n=\frac{1±79}{6} when ± is plus. Add 1 to 79.

$$n=\frac{80}{6}$$

Reduce the fraction \frac{80}{6} to lowest terms by extracting and canceling out 2.

$$n=\frac{40}{3}$$

Now solve the equation n=\frac{1±79}{6} when ± is minus. Subtract 79 from 1.

$$n=-\frac{78}{6}$$

Divide -78 by 6.

$$n=-13$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{40}{3} for x_{1} and -13 for x_{2}.

$$3n^{2}-n-520=3\left(n-\frac{40}{3}\right)\left(n-\left(-13\right)\right)$$

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

$$3n^{2}-n-520=3\left(n-\frac{40}{3}\right)\left(n+13\right)$$

Subtract \frac{40}{3} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3n^{2}-n-520=3\times \frac{3n-40}{3}\left(n+13\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3n^{2}-n-520=\left(3n-40\right)\left(n+13\right)$$

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 3

$$x ^ 2 -\frac{1}{3}x -\frac{520}{3} = 0$$

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

$$r + s = \frac{1}{3} rs = -\frac{520}{3}$$

Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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>

$$r = \frac{1}{6} - u s = \frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{520}{3}

$$(\frac{1}{6} - u) (\frac{1}{6} + u) = -\frac{520}{3}$$

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

$$\frac{1}{36} - u^2 = -\frac{520}{3}$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = -\frac{520}{3}-\frac{1}{36} = -\frac{6241}{36}$$

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

$$u^2 = \frac{6241}{36} u = \pm\sqrt{\frac{6241}{36}} = \pm \frac{79}{6}$$

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

$$r =\frac{1}{6} - \frac{79}{6} = -13 s = \frac{1}{6} + \frac{79}{6} = 13.333$$

Evaluate

$$\left(3n-40\right)\left(n+13\right)$$