Solve -a^2+12a-36

$$- a ^ { 2 } + 12 a - 36$$

Factor

$$-\left(a-6\right)^{2}$$

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.

$$p+q=12 pq=-\left(-36\right)=36$$

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 36.

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

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$p=6 q=6$$

Rewrite -a^{2}+12a-36 as \left(-a^{2}+6a\right)+\left(6a-36\right).

$$\left(-a^{2}+6a\right)+\left(6a-36\right)$$

Factor out -a in the first and 6 in the second group.

$$-a\left(a-6\right)+6\left(a-6\right)$$

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

$$\left(a-6\right)\left(-a+6\right)$$

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.

$$-a^{2}+12a-36=0$$

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.

$$a=\frac{-12±\sqrt{12^{2}-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-12±\sqrt{144+4\left(-36\right)}}{2\left(-1\right)}$$

Multiply 4 times -36.

$$a=\frac{-12±\sqrt{144-144}}{2\left(-1\right)}$$

Add 144 to -144.

$$a=\frac{-12±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$a=\frac{-12±0}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-12±0}{-2}$$

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 6 for x_{2}.

$$-a^{2}+12a-36=-\left(a-6\right)\left(a-6\right)$$

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.

$$x ^ 2 -12x +36 = 0$$

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

$$r + s = 12 rs = 36$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 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>

$$r = 6 - u s = 6 + u$$

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

$$(6 - u) (6 + u) = 36$$

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

$$36 - u^2 = 36$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 36-36 = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = 6$$

Evaluate

$$-\left(a-6\right)^{2}$$