Solve 3n^2-n-420

$$3 n ^ { 2 } - n - 420$$

Factor

$$\left(n-12\right)\left(3n+35\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3n^{2}+an+bn-420. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-420\right)=-1260$$

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 -1260.

$$1,-1260 2,-630 3,-420 4,-315 5,-252 6,-210 7,-180 9,-140 10,-126 12,-105 14,-90 15,-84 18,-70 20,-63 21,-60 28,-45 30,-42 35,-36$$

Calculate the sum for each pair.

$$1-1260=-1259 2-630=-628 3-420=-417 4-315=-311 5-252=-247 6-210=-204 7-180=-173 9-140=-131 10-126=-116 12-105=-93 14-90=-76 15-84=-69 18-70=-52 20-63=-43 21-60=-39 28-45=-17 30-42=-12 35-36=-1$$

The solution is the pair that gives sum -1.

$$a=-36 b=35$$

Rewrite 3n^{2}-n-420 as \left(3n^{2}-36n\right)+\left(35n-420\right).

$$\left(3n^{2}-36n\right)+\left(35n-420\right)$$

Factor out 3n in the first and 35 in the second group.

$$3n\left(n-12\right)+35\left(n-12\right)$$

Factor out common term n-12 by using distributive property.

$$\left(n-12\right)\left(3n+35\right)$$

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.

$$3n^{2}-n-420=0$$

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.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-420\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-1\right)±\sqrt{1-12\left(-420\right)}}{2\times 3}$$

Multiply -12 times -420.

$$n=\frac{-\left(-1\right)±\sqrt{1+5040}}{2\times 3}$$

Add 1 to 5040.

$$n=\frac{-\left(-1\right)±\sqrt{5041}}{2\times 3}$$

Take the square root of 5041.

$$n=\frac{-\left(-1\right)±71}{2\times 3}$$

The opposite of -1 is 1.

$$n=\frac{1±71}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{1±71}{6}$$

Now solve the equation n=\frac{1±71}{6} when ± is plus. Add 1 to 71.

$$n=\frac{72}{6}$$

Divide 72 by 6.

$$n=12$$

Now solve the equation n=\frac{1±71}{6} when ± is minus. Subtract 71 from 1.

$$n=-\frac{70}{6}$$

Reduce the fraction \frac{-70}{6} to lowest terms by extracting and canceling out 2.

$$n=-\frac{35}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 12 for x_{1} and -\frac{35}{3} for x_{2}.

$$3n^{2}-n-420=3\left(n-12\right)\left(n-\left(-\frac{35}{3}\right)\right)$$

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

$$3n^{2}-n-420=3\left(n-12\right)\left(n+\frac{35}{3}\right)$$

Add \frac{35}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3n^{2}-n-420=3\left(n-12\right)\times \frac{3n+35}{3}$$

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

$$3n^{2}-n-420=\left(n-12\right)\left(3n+35\right)$$

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

$$x ^ 2 -\frac{1}{3}x -140 = 0$$

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

$$r + s = \frac{1}{3} rs = -140$$

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 = -140

$$(\frac{1}{6} - u) (\frac{1}{6} + u) = -140$$

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

$$\frac{1}{36} - u^2 = -140$$

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

$$-u^2 = -140-\frac{1}{36} = -\frac{5041}{36}$$

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

$$u^2 = \frac{5041}{36} u = \pm\sqrt{\frac{5041}{36}} = \pm \frac{71}{6}$$

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

$$r =\frac{1}{6} - \frac{71}{6} = -11.667 s = \frac{1}{6} + \frac{71}{6} = 12$$

Evaluate

$$\left(n-12\right)\left(3n+35\right)$$