Answer:
To make the polynomial 9x^10 + ax^b + 100 a perfect square trinomial, we need to add a constant term to it such that it becomes a square of a binomial.
Let's first write the square of a binomial in general form:
(a + b)^2 = a^2 + 2ab + b^2
If we compare this general form with our polynomial, we can see that the first term, 9x^10, is equal to (3x^5)^2, which means that we can write our polynomial as:
(3x^5)^2 + ax^b + 100 = (3x^5 + c)^2
Expanding the right-hand side of this equation, we get:
(3x^5 + c)^2 = 9x^10 + 6cx^15 + c^2
Comparing the coefficient of x^15 on both sides, we get:
6c = 0
Since c cannot be zero (otherwise we would end up with the original polynomial), this means that we must have:
c = 0
Therefore, we can write our polynomial as:
(3x^5)^2 + ax^b + 100 = (3x^5)^2
Expanding the right-hand side, we get:
(3x^5)^2 = 9x^10
Therefore, we must have:
a = 0
b = 10
So the values of a and b that make the polynomial 9x^10 + ax^b + 100 a perfect square trinomial are a = 0 and b = 10.