Multiply x3 - 2x2 - 13x - 10
16a2b4
Given any number it is easy to find a rule based on a polynomial of order 5 such that the first five numbers are as listed in the question and the nth is the given number. There are also non-polynomial solutions.The simplest polynomial of smaller order is:Un = (n2 + n + 4)/2 for n = 1, 2, 3, ...
No integer roots. Quadratic formula gives 1.55 and -0.81 to the nearest hundredth.
take out zeros
by synthetic division and quadratic equation
Multiply x3 - 2x2 - 13x - 10
If the cubic polynomial you are given does not have an obvious factorization, then you must use synthetic division. I'm sure wikipedia can tell you all about that.
Find All Possible Roots/Zeros Using the Rational Roots Test f(x)=x^4-81 ... If a polynomial function has integer coefficients, then every rational zero will ...
If there is one variable. Then put each variable equal to zero and then solve for the other variable.
Do mean find the polynomial given its roots ? If so the answer is (x -r1)(x-r2)...(x-rn) where r1,r2,.. rn is the given list roots.
If you have the zeros of a polynomial, it is easy, almost trivial, to find an expression with those zeros. I am not sure I understood the question correctly, but let's assume you have the zero 2 with multiplicity 2, and other zeros at 3 and 5. Just write the expression: (x-2)(x-2)(x-3)(x-5). (Example with a negative zero: if there is a zero at "-5", the factor becomes (x- -5) = (x + 5).) You can multiply this out to get the polynomial if you like. For example, if you multiply every term in the first factor with every term in the second factor, you get x2 -2x -2x + 4 = x2 -4x + 4. Next, multiply each term of this polynomial with each term of the next factor, etc.
16a2b4
The remainder theorem states that if you divide a polynomial function by one of it's linier factors it's degree will be decreased by one. This theorem is often used to find the imaginary zeros of polynomial functions by reducing them to quadratics at which point they can be solved by using the quadratic formula.
If you know one linear factor, then divide the polynomial by that factor. The quotient will then be a polynomial whose order (or degree) is one fewer than that of the one that you stared with. The smaller order may make it easier to factorise.
The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.The answer can be any number that you like: it is always possible to find a polynomial of order 5 to fit the given numbers and any other number.The lowest degree polynomial that will fit the given numbers is the quadraticUn = (9n2 - 205n + 792)/2 for n = 1, 2, 3, .. . and that gives the next number as -57.
7w2 -17w+16 is a polynomial that cannot be factored. We call this a prime polynomial.There are also no like terms to combine. So nothing much more can be done with this polynomial. If you wanted to find the roots or the zeros, you could use the quadratic formula.