yes
No. Even if the answer is zero, zero is still a polynomial.
Let's try an example:The difference between (6x3 + x2 - 4x + 9) and (6x3 + x2 - 4x + 7) is 2 .2 is a polynomial of degree 0, so this example would appear to support the hypothesis in the question.However, polynomials cannot include negative exponents. So, (2x)/(2x2) produces 1/x, which is not a polynomial.So no, not always.
Equations will have an equals sign. Such as: x + 3 = 2 Polynomials will not. Such as: 2x + 3
A polynomial function is simply a function that is made of one or more mononomials. For example 4x^2+3x-5 A rational function is when a polynomial function is divided by another polynomial function.
We won't be able to answer this accurately without knowing the polynomials.
polynomials have 4 or more terms. I learned about that today in my math class. monomial =1 binomial=2 trinomial=3 polynomial=4+
The idea here is to multiply each term in the first polynomial by each term in the second polynomial.
A polynomial is any number of the form Ax^n + Bx^n-1 + ... + c. So, multiplying numbers with exponents with any other numbers with exponents in polynomials only results in another, larger polynomial. Since this is multiplication, you could call the resultant polynomial a product.
You can factor a polynomial using one of these steps: 1. Factor out the greatest common monomial factor. 2. Look for a difference of two squares or a perfect square trinomial. 3. Factor polynomials in the form ax^2+bx+c into a product of binomials. 4. Factor a polynomial with 4 terms by grouping.
A sum of polynomials is a polynomial.A product of polynomials is a polynomial.A composition of two polynomials is a polynomial, which is obtained by substituting a variable of the first polynomial by the second polynomial.The derivative of the polynomial anxn + an-1xn-1 + ... + a2x2 + a1x + a0 is the polynomial nanxn-1 + (n-1)an-1xn-2 + ... + 2a2x + a1. If the set of the coefficients does not contain the integers (for example if the coefficients are integers modulo some prime number p), then kak should be interpreted as the sum of ak with itself, k times. For example, over the integers modulo p, the derivative of the polynomial xp+1 is the polynomial 0.If the division by integers is allowed in the set of coefficients, a primitive or antiderivative of the polynomial anxn + an-1xn-1 + ... + a2x2 + a1x + a0 is anxn+1/(n+1) + an-1xn/n + ... + a2x3/3 + a1x2/2 + a0x +c, where c is an arbitrary constant. Thus x2+1 is a polynomial with integer coefficients whose primitives are not polynomials over the integers. If this polynomial is viewed as a polynomial over the integers modulo 3 it has no primitive at all.
The coefficients of polynomials are the numbers in front of the variable expressions. Ex: In the polynomial: 3x^5 + 12x^2 - 45x + 134 the numerical coefficients are: 3,12,& -45
That means that you divide one polynomial by another polynomial. Basically, if you have polynomials "A" and "B", you look for a polynomial "C" and a remainder "R", such that: B x C + R = A ... such that the remainder has a lower degree than polynomial "B", the polynomial by which you are dividing. For example, if you divide by a polynomial of degree 3, the remainder must be of degree 2 or less.