# Recently Added Math Formulas

Current Location  >  Math Formulas > Math Examples & Tutorials > Solving Second Degree Polynomials 2

# Solving Second Degree Polynomials 2

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

Solution:

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

Therefore, the factors for the given second degree polynomial equation x2-44x+ 435 = 0 are (x -29) and (x- 15).

Example 2: Find the roots of 3 x2 + x + 6.

Solution:

Factoring doesn't work (trust me!), so use the quadratic formula: The roots are

[-1 ± sqrt( 12 - 4.3.6 )] / 2.3
= (-1 ± sqrt (-71)) / 6
= not a real number.

Therefore, 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.

Solution:

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.
i. e. x  – x – 6 = 0.

Web-Formulas.com © 2019 | Contact us | Terms of Use | Privacy Policy |  