-2
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
The square root of a polynomial is another polynomial that, when multiplied by itself, yields the original polynomial. Not all polynomials have a square root that is also a polynomial; for example, the polynomial (x^2 + 1) does not have a polynomial square root in the real number system. However, some polynomials, like (x^2 - 4), have polynomial square roots, which in this case would be (x - 2) and (x + 2). Finding the square root of a polynomial can involve techniques such as factoring or using the quadratic formula for quadratic polynomials.
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.
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) ).
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.