Solve a^2-2a-3

$$a ^ { 2 } - 2 a - 3$$

Factor

$$\left(a-3\right)\left(a+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-3. To find p and q, set up a system to be solved.

$$p+q=-2 pq=1\left(-3\right)=-3$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$p=-3 q=1$$

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

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

Factor out a in a^{2}-3a.

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

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

$$\left(a-3\right)\left(a+1\right)$$

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.

$$a^{2}-2a-3=0$$

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.

$$a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)}}{2}$$

Square -2.

$$a=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

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

Add 4 to 12.

$$a=\frac{-\left(-2\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$a=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$a=\frac{2±4}{2}$$

Now solve the equation a=\frac{2±4}{2} when ± is plus. Add 2 to 4.

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

Divide 6 by 2.

$$a=3$$

Now solve the equation a=\frac{2±4}{2} when ± is minus. Subtract 4 from 2.

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

Divide -2 by 2.

$$a=-1$$

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 -1 for x_{2}.

$$a^{2}-2a-3=\left(a-3\right)\left(a-\left(-1\right)\right)$$

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

$$a^{2}-2a-3=\left(a-3\right)\left(a+1\right)$$

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.

$$x ^ 2 -2x -3 = 0$$

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

$$r + s = 2 rs = -3$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(1 - u) (1 + u) = -3$$

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

$$1 - u^2 = -3$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -3-1 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =1 - 2 = -1 s = 1 + 2 = 3$$

Evaluate

$$\left(a-3\right)\left(a+1\right)$$