No. A polynomial has positive powers of the variable.
The polynomial is (x + 1)*(x + 1)*(x - 1) = x3 + x2 - x - 1
A zero of a polynomial function - or of any function, for that matter - is a value of the independent variable (often called "x") for which the function evaluates to zero. In other words, a solution to the equation P(x) = 0. For example, if your polynomial is x2 - x, the corresponding equation is x2 - x = 0. Solutions to this equation - and thus, zeros to the polynomial - are x = 0, and x = 1.
You need to multiply three terms, one for each zero. To have only two zeros, the polynomial would need to have a "double zero" (or more generally, a "multiple zero), that is, a repeated factor. In this case, the zeros can be one of the following: -1, -1, 1, with the corresponding factors: (x+1)(x+1)(x-1) or: -1, 1, 1, with the corresponding factors: (x+1)(x-1)(x-1) If you like, you can multiply these factors out to get the polynomial in standard form.
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A polynomial is an algebraic expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication operations. It typically has one or more terms, each of which can have a different degree. A multinomial, on the other hand, is a specific type of polynomial that has more than one term with multiple variables raised to different powers. In essence, all multinomials are polynomials, but not all polynomials are multinomials.
what kind of polynomial is shown 3x3+x+1
The polynomial is (x + 1)*(x + 1)*(x - 1) = x3 + x2 - x - 1
The polynomial P(x)=(x-3)(x-0)(x+3)(x-1) is of the fourth degree.
A prime polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials over its coefficient field. In other words, it has no divisors other than itself and the unit (constant) polynomials. For example, in the field of real numbers, (x^2 + 1) is a prime polynomial because it cannot be factored into real linear factors. Conversely, polynomials like (x^2 - 1) are not prime because they can be factored as ((x - 1)(x + 1)).
X2 - X - 2(X + 1)(X - 2)===============(X + 1) is a factor of the above polynomial.
Yes, it is.
(x + 1)(x - 1)
Polynomials are not closed under division because dividing one polynomial by another can result in a quotient that is not a polynomial. Specifically, when a polynomial is divided by another polynomial of a higher degree, the result can be a rational function, which includes terms with variables in the denominator. For example, dividing (x^2) by (x) gives (x), a polynomial, but dividing (x) by (x^2) results in (\frac{1}{x}), which is not a polynomial. Thus, the closure property does not hold for polynomial division.
The polynomial equation is x2 - x - 1 = 0.
(x + 8)(x + 1)
(x + 8)(x + 1)
Yes, that's correct. According to the Factor Theorem, if a polynomial ( P(x) ) is divided by ( (x - a) ) and the remainder is zero, then ( (x - a) ) is indeed a factor of the polynomial. This means that ( P(a) = 0 ), indicating that ( a ) is a root of the polynomial. Thus, the polynomial can be expressed as ( P(x) = (x - a)Q(x) ) for some polynomial ( Q(x) ).