Solve n^2+n-42

$$n ^ { 2 } + n - 42$$

Factor

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

Solution Steps

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

$$a+b=1 ab=1\left(-42\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

Rewrite n^{2}+n-42 as \left(n^{2}-6n\right)+\left(7n-42\right).

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

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

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

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

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

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.

$$n^{2}+n-42=0$$

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.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-42\right)}}{2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\left(-42\right)}}{2}$$

Multiply -4 times -42.

$$n=\frac{-1±\sqrt{1+168}}{2}$$

Add 1 to 168.

$$n=\frac{-1±\sqrt{169}}{2}$$

Take the square root of 169.

$$n=\frac{-1±13}{2}$$

Now solve the equation n=\frac{-1±13}{2} when ± is plus. Add -1 to 13.

$$n=\frac{12}{2}$$

Divide 12 by 2.

$$n=6$$

Now solve the equation n=\frac{-1±13}{2} when ± is minus. Subtract 13 from -1.

$$n=-\frac{14}{2}$$

Divide -14 by 2.

$$n=-7$$

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 -7 for x_{2}.

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

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

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

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.

$$x ^ 2 +1x -42 = 0$$

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

$$r + s = -1 rs = -42$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -42

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -42$$

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

$$\frac{1}{4} - u^2 = -42$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -42-\frac{1}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =-\frac{1}{2} - \frac{13}{2} = -7 s = -\frac{1}{2} + \frac{13}{2} = 6$$

Evaluate

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