Solve -a^2+a+6=

$$- a ^ { 2 } + a + 6 =$$

Factor

$$-\left(a-3\right)\left(a+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa+6. To find p and q, set up a system to be solved.

$$p+q=1 pq=-6=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$p=3 q=-2$$

Rewrite -a^{2}+a+6 as \left(-a^{2}+3a\right)+\left(-2a+6\right).

$$\left(-a^{2}+3a\right)+\left(-2a+6\right)$$

Factor out -a in the first and -2 in the second group.

$$-a\left(a-3\right)-2\left(a-3\right)$$

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

$$\left(a-3\right)\left(-a-2\right)$$

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.

$$-a^{2}+a+6=0$$

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.

$$a=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-1±\sqrt{1+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$a=\frac{-1±\sqrt{1+24}}{2\left(-1\right)}$$

Add 1 to 24.

$$a=\frac{-1±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$a=\frac{-1±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-1±5}{-2}$$

Now solve the equation a=\frac{-1±5}{-2} when ± is plus. Add -1 to 5.

$$a=\frac{4}{-2}$$

Divide 4 by -2.

$$a=-2$$

Now solve the equation a=\frac{-1±5}{-2} when ± is minus. Subtract 5 from -1.

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

Divide -6 by -2.

$$a=3$$

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

$$-a^{2}+a+6=-\left(a-\left(-2\right)\right)\left(a-3\right)$$

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

$$-a^{2}+a+6=-\left(a+2\right)\left(a-3\right)$$

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.

$$x ^ 2 -1x -6 = 0$$

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

$$r + s = 1 rs = -6$$

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 = -6

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -6$$

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

$$\frac{1}{4} - u^2 = -6$$

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

$$-u^2 = -6-\frac{1}{4} = -\frac{25}{4}$$

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

$$u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{2}$$

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

$$r =\frac{1}{2} - \frac{5}{2} = -2 s = \frac{1}{2} + \frac{5}{2} = 3$$

Evaluate

$$-\left(a-3\right)\left(a+2\right)$$