Solve n^2+15n+54

$$n ^ { 2 } + 15 n + 54$$

Factor

$$\left(n+6\right)\left(n+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+54. To find a and b, set up a system to be solved.

$$a+b=15 ab=1\times 54=54$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 54.

$$1,54 2,27 3,18 6,9$$

Calculate the sum for each pair.

$$1+54=55 2+27=29 3+18=21 6+9=15$$

The solution is the pair that gives sum 15.

$$a=6 b=9$$

Rewrite n^{2}+15n+54 as \left(n^{2}+6n\right)+\left(9n+54\right).

$$\left(n^{2}+6n\right)+\left(9n+54\right)$$

Factor out n in the first and 9 in the second group.

$$n\left(n+6\right)+9\left(n+6\right)$$

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

$$\left(n+6\right)\left(n+9\right)$$

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.

$$n^{2}+15n+54=0$$

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.

$$n=\frac{-15±\sqrt{15^{2}-4\times 54}}{2}$$

Square 15.

$$n=\frac{-15±\sqrt{225-4\times 54}}{2}$$

Multiply -4 times 54.

$$n=\frac{-15±\sqrt{225-216}}{2}$$

Add 225 to -216.

$$n=\frac{-15±\sqrt{9}}{2}$$

Take the square root of 9.

$$n=\frac{-15±3}{2}$$

Now solve the equation n=\frac{-15±3}{2} when ± is plus. Add -15 to 3.

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

Divide -12 by 2.

$$n=-6$$

Now solve the equation n=\frac{-15±3}{2} when ± is minus. Subtract 3 from -15.

$$n=-\frac{18}{2}$$

Divide -18 by 2.

$$n=-9$$

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 -9 for x_{2}.

$$n^{2}+15n+54=\left(n-\left(-6\right)\right)\left(n-\left(-9\right)\right)$$

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

$$n^{2}+15n+54=\left(n+6\right)\left(n+9\right)$$

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.

$$x ^ 2 +15x +54 = 0$$

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

$$r + s = -15 rs = 54$$

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 = 54

$$(-\frac{15}{2} - u) (-\frac{15}{2} + u) = 54$$

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

$$\frac{225}{4} - u^2 = 54$$

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

$$-u^2 = 54-\frac{225}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =-\frac{15}{2} - \frac{3}{2} = -9 s = -\frac{15}{2} + \frac{3}{2} = -6$$

Evaluate

$$\left(n+6\right)\left(n+9\right)$$