Solve a^2+9a-36

$$a ^ { 2 } + 9 a - 36$$

Factor

$$\left(a-3\right)\left(a+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-36. To find p and q, set up a system to be solved.

$$p+q=9 pq=1\left(-36\right)=-36$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$p=-3 q=12$$

Rewrite a^{2}+9a-36 as \left(a^{2}-3a\right)+\left(12a-36\right).

$$\left(a^{2}-3a\right)+\left(12a-36\right)$$

Factor out a in the first and 12 in the second group.

$$a\left(a-3\right)+12\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(a+12\right)$$

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.

$$a^{2}+9a-36=0$$

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.

$$a=\frac{-9±\sqrt{9^{2}-4\left(-36\right)}}{2}$$

Square 9.

$$a=\frac{-9±\sqrt{81-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$a=\frac{-9±\sqrt{81+144}}{2}$$

Add 81 to 144.

$$a=\frac{-9±\sqrt{225}}{2}$$

Take the square root of 225.

$$a=\frac{-9±15}{2}$$

Now solve the equation a=\frac{-9±15}{2} when ± is plus. Add -9 to 15.

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

Now solve the equation a=\frac{-9±15}{2} when ± is minus. Subtract 15 from -9.

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

Divide -24 by 2.

$$a=-12$$

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

$$a^{2}+9a-36=\left(a-3\right)\left(a-\left(-12\right)\right)$$

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

$$a^{2}+9a-36=\left(a-3\right)\left(a+12\right)$$

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.

$$x ^ 2 +9x -36 = 0$$

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

$$r + s = -9 rs = -36$$

Two numbers r and s sum up to -9 exactly when the average of the two numbers is \frac{1}{2}*-9 = -\frac{9}{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>

$$r = -\frac{9}{2} - u s = -\frac{9}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -36

$$(-\frac{9}{2} - u) (-\frac{9}{2} + u) = -36$$

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

$$\frac{81}{4} - u^2 = -36$$

Simplify the expression by subtracting \frac{81}{4} on both sides

$$-u^2 = -36-\frac{81}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =-\frac{9}{2} - \frac{15}{2} = -12 s = -\frac{9}{2} + \frac{15}{2} = 3$$

Evaluate

$$\left(a-3\right)\left(a+12\right)$$