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# Polynomials Second Degree

A second degree polynomial is generally expressed as below:

P(x) = a ∙ x2 + b ∙ x2 + c, and a ≠ 0

P(x) can also be rewritten as: a(x - x1)(x - x2)

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

x1 + x2 = - b/a
x1 ∙ x2 = c/a

x1 and x2 are the possible solutions for P(x)

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

x1 = (-b + √(b2 - 4ac)) / 2a
x2 = (-b - √(b2 - 4ac)) / 2a

b2 - 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:

x1 = x2, if b2 - 4ac = 0, there exists only 1 solution
x1 ≠ x2, if b2 - 4ac > 0, there exists 2 solutions
there exists no solutions if b2 - 4ac < 0

Example 1: Predict the factors for the second degree polynomial equation x2-44x+ 435 = 0.

The given second degree polynomial equation is x2-44x+ 435 = 0.

Step 1: x2-44x+ 435 = x2- 29x- 15x+ (-29 x -15)

Step 2: x2-44x+ 435 = x(x- 29) - 15(x- 29)

Step 3: x2-44x+ 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 x2-44x+ 435 = 0 are therefore (x -29) and (x- 15).

Example 2: Find the roots of 3 x2 + 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 b2 - 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
x2 – (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.

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