Solve 3n^2-11n-42

$$3 n ^ { 2 } - 11 n - 42$$

Factor

$$\left(n-6\right)\left(3n+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3n^{2}+an+bn-42. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\left(-42\right)=-126$$

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 -126.

$$1,-126 2,-63 3,-42 6,-21 7,-18 9,-14$$

Calculate the sum for each pair.

$$1-126=-125 2-63=-61 3-42=-39 6-21=-15 7-18=-11 9-14=-5$$

The solution is the pair that gives sum -11.

$$a=-18 b=7$$

Rewrite 3n^{2}-11n-42 as \left(3n^{2}-18n\right)+\left(7n-42\right).

$$\left(3n^{2}-18n\right)+\left(7n-42\right)$$

Factor out 3n in the first and 7 in the second group.

$$3n\left(n-6\right)+7\left(n-6\right)$$

Factor out common term n-6 by using distributive property.

$$\left(n-6\right)\left(3n+7\right)$$

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.

$$3n^{2}-11n-42=0$$

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.

$$n=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\left(-42\right)}}{2\times 3}$$

Square -11.

$$n=\frac{-\left(-11\right)±\sqrt{121-4\times 3\left(-42\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-11\right)±\sqrt{121-12\left(-42\right)}}{2\times 3}$$

Multiply -12 times -42.

$$n=\frac{-\left(-11\right)±\sqrt{121+504}}{2\times 3}$$

Add 121 to 504.

$$n=\frac{-\left(-11\right)±\sqrt{625}}{2\times 3}$$

Take the square root of 625.

$$n=\frac{-\left(-11\right)±25}{2\times 3}$$

The opposite of -11 is 11.

$$n=\frac{11±25}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{11±25}{6}$$

Now solve the equation n=\frac{11±25}{6} when ± is plus. Add 11 to 25.

$$n=\frac{36}{6}$$

Divide 36 by 6.

$$n=6$$

Now solve the equation n=\frac{11±25}{6} when ± is minus. Subtract 25 from 11.

$$n=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$n=-\frac{7}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{7}{3} for x_{2}.

$$3n^{2}-11n-42=3\left(n-6\right)\left(n-\left(-\frac{7}{3}\right)\right)$$

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

$$3n^{2}-11n-42=3\left(n-6\right)\left(n+\frac{7}{3}\right)$$

Add \frac{7}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3n^{2}-11n-42=3\left(n-6\right)\times \frac{3n+7}{3}$$

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

$$3n^{2}-11n-42=\left(n-6\right)\left(3n+7\right)$$

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 3

$$x ^ 2 -\frac{11}{3}x -14 = 0$$

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

$$r + s = \frac{11}{3} rs = -14$$

Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{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>

$$r = \frac{11}{6} - u s = \frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -14

$$(\frac{11}{6} - u) (\frac{11}{6} + u) = -14$$

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

$$\frac{121}{36} - u^2 = -14$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = -14-\frac{121}{36} = -\frac{625}{36}$$

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

$$u^2 = \frac{625}{36} u = \pm\sqrt{\frac{625}{36}} = \pm \frac{25}{6}$$

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

$$r =\frac{11}{6} - \frac{25}{6} = -2.333 s = \frac{11}{6} + \frac{25}{6} = 6$$

Evaluate

$$\left(n-6\right)\left(3n+7\right)$$