The following polynomial is prime. 2q² + 9q + 1
If prime ----> can't be factored over the rational numbers
no rational factors if its (b^2 - 4ac) is NOT a perfect square.
b^2 - 4ac = 81 - 4(2)(1) = 73
which is not a perfect square , so ....
end of discussion.
What is the question, anna? Do we need to factor it?
To determine whether the polynomial 2q² + 9q + 1 is prime, we need to check if it can be factored into smaller terms.
In this case, we can use the quadratic formula to determine the roots of the polynomial. The quadratic formula is written as:
q = [-b ± √(b² - 4ac)] / 2a
For our polynomial 2q² + 9q + 1, a = 2, b = 9, and c = 1. Plugging these values into the quadratic formula, we get:
q = [-9 ± √(9² - 4*2*1)] / (2*2)
q = [-9 ± √(81 - 8)] / 4
q = [-9 ± √73] / 4
Since the discriminant (the value inside the square root) is 73, which cannot be factored into smaller terms, we can conclude that 2q² + 9q + 1 is indeed a prime polynomial.