Solve x^2-2x-3-0

$$x ^ { 2 } - 2 x - 3 - 0$$

Evaluate

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

Solution Steps

Multiply -1 and 0 to get 0.

$$x^{2}-2x-3+0$$

Add -3 and 0 to get -3.

$$x^{2}-2x-3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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

$$a=-3 b=1$$

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

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

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

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

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

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

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.

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

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.

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

Square -2.

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

Multiply -4 times -3.

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

Add 4 to 12.

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

Take the square root of 16.

$$x=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±4}{2}$$

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

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

Divide 6 by 2.

$$x=3$$

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

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

Divide -2 by 2.

$$x=-1$$

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}.

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

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

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