The only thing in the question I am not totally certain about is the definition of a 'zero' on a polynomial - I will assume this means an x-intercept, which seems to make sense since these are the points where the value of the function is zero.
Taking the first polynomial: the maximum number of turning points for a polynomial of order n is (n-1). Take the example of a quadratic, which always has 1 turning point. Therefore the minimum order of the first polynomial is 6.
The maximum number of x-intercepts for a polynomial of order n is n. Therefore the second polynomial has a minimum order of 6.
Multiplication of two polynomials can get very messy very quickly. However, picture putting the two in brackets next to each other, such that the terms are in decreasing orders. You can easily see the maximum order term is found by multiplying the first term in each bracket. A polynomial's order is judged solely on the maximum power of the variable, so this is all we need to consider.
This multiplication becomes ax^6 * bx^6 where a and b are arbitrary constants in this context. Hence this product is abx^12 (exponents add when the terms are multiplied, where 12 = 6 + 6), so minimum degree of new polynomial = 12
Answer:
4^8
Step-by-step explanation:
If the second 4 is an exponent, as in (4^2)^4, then multiply the exponents.
(4^2)^4 = 4^(2 * 4) = 4^8
Hi there!
We can use the discriminant of the quadratic formula to find the number of real solutions to this (or any other quadratic) equation.
The discriminant can be found by plugging in the data from the equation into the following formula:
When we fill in the data from the question we get
Since d < 0 the equation has zero real solutions.
