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Q: What Must the sum of three polynomials again be a polynomial?
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Why is it important to know the reverse process of multiplication?

To cross-check that a multiplication is correct as for example if 7*8 = 56 then the reverse process of division must be correct as 56/7 = 8 or 56/8 = 7


How can you tell if a polynomial is a perfect square?

Let's take a quadratic polynomial. There are three terms in a quadratic polynomial. Example: X^2 + 8X + 16 = 0 To satisfy the criteria of a perfect square polynomial, the first and last term of the polynomial must be squares. The middle term must be either plus or minus two multiplied by the square root of the first term multiplied by the square root of the last term. If these three criteria are satisifed, the polynomial is a perfect square. Let us take the above quadratic. X^2 + 8X + 16 = X^2 + 2(4X) + 4^2 = (X+4)^2 As we can see, each criteria is satified and the polynomial does indeed form a perfect square.


What is a quadratic polynomial which has no zeros?

A quadratic polynomial must have zeros, though they may be complex numbers.A quadratic polynomial with no real zeros is one whose discriminant b2-4ac is negative. Such a polynomial has no special name.


Can a polynomial equation have no answer?

It depends on the domain. In the complex domain, a polynomial of order n must have n solutions, although some of these may be multiple solutions. In the real domain, a polynomial of odd order must have at least one real solution, while a polynomial of even order may have no real solutions.


Why the graph of a polynomial function with real coefficients must have a y-intercept but may have no x-intercept?

For a polynomial of the form y = p(x) (i.e., some polynomial function of x), having a y-intercept simply means that the polynomial is defined for x = 0 - and a polynomial is defined for any value of "x". As for the x-intercept: from left to right, a polynomial of even degree may come down, not quite reach zero, and then go back up again. A simple example is y = x2 + 1. Why is the situation for "x" and for "y" different? Well, the original equation is a polynomial in "x"; but if you solve for "x", you don't get a polynomial in "y".