Solve a^2+27a+180

$$a ^ { 2 } + 27 a + 180$$

Factor

$$\left(a+12\right)\left(a+15\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+180. To find p and q, set up a system to be solved.

$$p+q=27 pq=1\times 180=180$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 27.

$$p=12 q=15$$

Rewrite a^{2}+27a+180 as \left(a^{2}+12a\right)+\left(15a+180\right).

$$\left(a^{2}+12a\right)+\left(15a+180\right)$$

Factor out a in the first and 15 in the second group.

$$a\left(a+12\right)+15\left(a+12\right)$$

Factor out common term a+12 by using distributive property.

$$\left(a+12\right)\left(a+15\right)$$

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.

$$a^{2}+27a+180=0$$

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.

$$a=\frac{-27±\sqrt{27^{2}-4\times 180}}{2}$$

Square 27.

$$a=\frac{-27±\sqrt{729-4\times 180}}{2}$$

Multiply -4 times 180.

$$a=\frac{-27±\sqrt{729-720}}{2}$$

Add 729 to -720.

$$a=\frac{-27±\sqrt{9}}{2}$$

Take the square root of 9.

$$a=\frac{-27±3}{2}$$

Now solve the equation a=\frac{-27±3}{2} when ± is plus. Add -27 to 3.

$$a=-\frac{24}{2}$$

Divide -24 by 2.

$$a=-12$$

Now solve the equation a=\frac{-27±3}{2} when ± is minus. Subtract 3 from -27.

$$a=-\frac{30}{2}$$

Divide -30 by 2.

$$a=-15$$

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 -15 for x_{2}.

$$a^{2}+27a+180=\left(a-\left(-12\right)\right)\left(a-\left(-15\right)\right)$$

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

$$a^{2}+27a+180=\left(a+12\right)\left(a+15\right)$$

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.

$$x ^ 2 +27x +180 = 0$$

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

$$r + s = -27 rs = 180$$

Two numbers r and s sum up to -27 exactly when the average of the two numbers is \frac{1}{2}*-27 = -\frac{27}{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>

$$r = -\frac{27}{2} - u s = -\frac{27}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 180

$$(-\frac{27}{2} - u) (-\frac{27}{2} + u) = 180$$

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

$$\frac{729}{4} - u^2 = 180$$

Simplify the expression by subtracting \frac{729}{4} on both sides

$$-u^2 = 180-\frac{729}{4} = -\frac{9}{4}$$

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.

$$r =-\frac{27}{2} - \frac{3}{2} = -15 s = -\frac{27}{2} + \frac{3}{2} = -12$$

Evaluate

$$\left(a+12\right)\left(a+15\right)$$