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How to tell if there are no real roots?
The real roots of what, exactly? If you mean a square trinomial, then: If the discriminant is positive, the polynomial has two real roots. If the discriminant is zero, the polynomial has one (double) real root. If the discriminant is negative, the polynomial has two complex roots (and of course no real roots). The discriminant is the term under the square root in the quadratic equation, in other words, b2 - 4ac.
What is the polynomial having two terms?
A polynomial with two terms is called a binomial.
Will the product of two polynomials always be a polynomial?
Yes, the product of two polynomials will always be a polynomial. This is because when you multiply two polynomials, you are essentially combining like terms and following the rules of polynomial multiplication, which results in a new polynomial with coefficients that are the products of the corresponding terms in the original polynomials. Therefore, the product of two polynomials will always be a polynomial.
What are the different forms in quadratic equation?
The general form of a quadratic equation is y = ax2 + bx + c, where a, b and c are real constants and a ≠0. If a = 0 then it is not a quadratic! There are two ways of classifying the forms. "CUP OR CAP" If a > 0 then the graph of the quadratic is cup shaped - like a U. If a < 0 then the graph is cap shaped - like an inverted U. "NUMBER OF ROOTS" Using the above form, calculate the discriminant, d = b2 - 4ac If d > 0 the quadratic has two real roots. That is, two distinct real values of x for which y = 0. If d = 0 the quadratic has two coincident real roots. (Some consider this as one root but it is useful to consider the situation as two roots that coincide since that approach maintains parity between the number of roots and the order of the polynomial.) If d < 0 there are no real roots. Instead, it has two complex roots which will be conjugates of one another.
What is x to the second power plus x to the fifth power?
It is a fifth order polynomial. The two terms cannot be combined, except to factor out x² and get x²(x³ + 1). This can be solved for 5 roots: 0, 0, -1, and two complex roots: 1/2 ± i(√3)/2
