Q: What is a polynomial with a single root at x equals -6 and a double root at x equals -4?

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That would be (x - 2) ( x - 5) ( x - 5). If you like, you can multiply these polynomials to get a single polynomial in standard form (i.e., not factored).

-2

If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.

The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1

A "root" of a polynomial is any value which, when replaced for the variable, results in the polynomial evaluating to zero. For example, in the polynomial x2 - 9, if you replace "x" by 3, or by -3, the resulting expression is equal to zero.

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That would be (x - 2) ( x - 5) ( x - 5). If you like, you can multiply these polynomials to get a single polynomial in standard form (i.e., not factored).

-2

4-17i

A root.

There are infinitely many polynomials which meet the requirement.The polynomial is x2 + (-a - bi - 3)x + (2a - 2bi - ai + b) = 0and the other root is x = a + bi.There is nothing in the question which requires its coefficients to be real.

1+x2 is a polynomial and doesn't have a real root.

There is not enough information. You can't calculate one root on the basis of another root. HOWEVER, if we assume that all the polynomial's coefficients are real, then if the polynomial has a complex root, then the complex conjugate of that root (in this case, 4 - 17i) must also be a root.

If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.

The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1

Yes, it is a linear polynomial.

No,

It is a polynomial if the square root is in a coefficient but not if it is applied to the variable. A polynomial can have only integer powers of the variable. Thus: sqrt(2)*x3 + 4*x + 3 is a polynomial expression but 2*x3 + 4*sqrt(x) + 3 is not.