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Which is a third degree polynomial with -1 and 1 as its only zeros?
Wiki User
∙ 9y ago
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.
Wiki User
∙ 9y ago
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What is the polynomial degree of 4ab5 2ab-3a4b3?
To find the polynomial degree you have only to add the exponents of all of the different components of the polynomial. In your case, you would add 1 and 5 from 4ab5 to get 6, 1 and 1 from 2ab to get 2, and 4 and 3 from 3a4b3 to get 7. Since the degree of the third component is the highest, that is you're answer.
Is it possible for a polynomial function of degree 3 to have no real zeros?
Yes - but only if the domain is restricted. Normally the domain is the whole of the real numbers and over that domain it must have at least one real zero.
What are the steps to solve these problems Find the degree of each polynomial 5a-2b2 plus 1 and 24xy-xy3 plus x2?
The degree of a polynomial function is the highest power any single term is raised to. For example, (5a - 2b^2) is a second degree function because the "b^2" is raised to the second power and the "a" is only raised to the (implied) first power. For (24xy-xy^3 + x^2) it is a third degree polynomial because the highest power is the cube of -xy.
What is the number in front of the term with the highest degree in a polynomial?
It is the Coefficient. It only refers to the given term that it is front. e.g. 2x^2 - 3x + 1 The '2' in front of 'x^2' only refers to 'x^2'. The '-3' in front of 'x' is the coefficient of '-3' The '1' is a constant.
Is it possible to find a polynomial of degree 3 that has -2 as its only real zero?
5
Are there only 3 degree's in a polynomial equation?
No. A polynomial can have as many degrees as you like.
What is the polynomial degree of 4ab5 2ab-3a4b3?
To find the polynomial degree you have only to add the exponents of all of the different components of the polynomial. In your case, you would add 1 and 5 from 4ab5 to get 6, 1 and 1 from 2ab to get 2, and 4 and 3 from 3a4b3 to get 7. Since the degree of the third component is the highest, that is you're answer.
What is the factor theorem?
In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.The factor theorem states that a polynomial has a factor if and only if
What does factor theorem mean?
In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.The factor theorem states that a polynomial has a factor if and only if
Is it possible for a polynomial function of degree 3 to have no real zeros?
Yes - but only if the domain is restricted. Normally the domain is the whole of the real numbers and over that domain it must have at least one real zero.
What is a polynomial with a degree of 3 and zeros of -3 and 3-2i?
The question has a simple answer only if the polynomial has rational coefficients. However, the question does not state that it has rational coefficients so it is not valid to assume that is the case. So, suppose the third root is p. Then the polynomial is (x + 3)*(x - 3 + 2i)*(x - p) = (x2 + 2xi - 9 + 6i)*(x - p) = x3 + (2i - p)x2 - (2pi - 6i + 9)x + (9p + 6pi)
What are the steps to solve these problems Find the degree of each polynomial 5a-2b2 plus 1 and 24xy-xy3 plus x2?
The degree of a polynomial function is the highest power any single term is raised to. For example, (5a - 2b^2) is a second degree function because the "b^2" is raised to the second power and the "a" is only raised to the (implied) first power. For (24xy-xy^3 + x^2) it is a third degree polynomial because the highest power is the cube of -xy.
What is the number in front of the term with the highest degree in a polynomial?
It is the Coefficient. It only refers to the given term that it is front. e.g. 2x^2 - 3x + 1 The '2' in front of 'x^2' only refers to 'x^2'. The '-3' in front of 'x' is the coefficient of '-3' The '1' is a constant.
Is it possible to find a polynomial of degree 3 that has -2 as its only real zero?
5
Is it always true that between any two zeros of the derivative of any polynomial there is a zero of the polynomial?
No. Consider the polynomial: f(x) = x3 + 4x2 + 4x + 16 then f'(x) = 3x2 + 8x + 4 = (3x + 2)(x + 2) => x = -2/3, -2 are the zeros of f'(x) Using the second derivative: f''(x) = 6x + 8 it can be seen that: f''(-2) = -4 -> x = -2 is a maximum f''(-2/3) = +4 -> x = -2/3 is a minimum But plugging back into the original polynomial: f(-2) = 16 f(-2/3) = 14 22/27 Between the zeros of the first derivative, the slope of the polynomial is negative so that the polynomial is always decreasing in value, but as the polynomial is greater than zero at the zeros of the first derivative, it cannot become zero between them. That is it has no zeros between the zeros of its first derivative f(x) = x3 + 4x2 + 4x + 16 = (x + 4)(x2 + 4) has only 1 zero at x = -4.
The polynomial given below has rootss?
You forgot to copy the polynomial. However, the Fundamental Theorem of Algebra states that every polynomial has at least one root, if complex roots are allowed. If a polynomial has only real coefficients, and it it of odd degree, it will also have at least one real solution.
Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
