A second degree polynomial is generally expressed as below:

*P*_{(x)} = a ∙ **x**^{2} + b ∙ **x**^{2} + c, and *a ≠ 0*

*P*_{(x)} can also be rewritten as: a(**x** - x_{1})(**x** - x_{2})

For any second degree polynomial that satisfies the conditions above we have:

x_{1} + x_{2} = - b/a

x_{1} ∙ x_{2} = c/a

x_{1} and x_{2} are the possible solutions for *P*_{(x)}

The solutions of a second degree can be easily calculated using the quadratic formulas shown below:

x_{1} = (-b + √(b^{2} - 4ac)) / 2a

x2 = (-b - √(b^{2} - 4ac)) / 2a

b^{2} - 4ac is called the discriminant of the quadratic formula. By analyzing the discriminant it is possible to find out how many solutions *P*(**x**) has:

x_{1} = x_{2}, if b^{2} - 4ac = 0, there exists only 1 solution

x_{1} ≠ x_{2}, if b^{2} - 4ac > 0, there exists 2 solutions

there exists no solutions if b^{2} - 4ac < 0

**Example 1: **Predict the factors for the second degree polynomial equation *x*^{2}-44x+ 435 = 0.

The given second degree polynomial equation is *x*^{2}-44x+ 435 = 0.** **

Step 1: *x*^{2}-44x+ 435 = x^{2}- 29x- 15x+ (-29 x -15)

Step 2: *x*^{2}-44x+ 435 = x(x- 29) - 15(x- 29)

Step 3: *x*^{2}-44x+ 435 = (x- 29) (x- 15)

Step 4: *x- 29 = 0 and x-15 = 0 *

Step 5: *(x- 29) (x-15) = 0 *

The factors for the given second degree polynomial equation x^{2}-44x+ 435 = 0 are therefore (x -29) and (x- 15).

**Example 2: **Find the roots of *3 x*^{2} + x + 6.

In this example we will use the quadratic formula to determine its roots, where we have:

a = 3

b = 1

c = 6

since b^{2} - 4ac = 1 – 4*3*6 = -71 < 0, we conclude that the polynomial has no real roots but there are two complex roots, namely x = ( -1 + sqrt(71)i ) / 6 and x = ( -1 + sqrt(71)i ) / 6.

**Example 3:** Find the quadratic equation whose roots are 3, -2.

The given roots are 3, -2.

Sum of the roots = 3 + (-2) = 3 – 2 = 1;

Product of the roots = 3 x (-2) = -6.

We know the Quadratic Equation whose roots are given is

x^{2} – (sum of the roots) x + (product of the roots) = 0.

So, the required equation is *x – (1) x + (-6) = 0*.

ie. *x – x – 6 = 0*.