Example 1: Predict the factors for the second degree polynomial equation x^{2}-44x+ 435 = 0.
Solution:
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
Therefore, the factors for the given second degree polynomial equation x^{2}-44x+ 435 = 0 are (x -29) and (x- 15).
Example 2: Find the roots of 3 x^{2} + x + 6.
Solution:
Factoring doesn't work (trust me!), so use the quadratic formula: The roots are
[-1 ± sqrt( 1^{2} - 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
x^{2} – (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.