4
A rational equation can be multiplied by the least common multiple of its denominators to make it into a polynomial equation. The degree of this polynomial is the highest power (of the variable) that appears in it. It can be proven that a polynomial of degree n must have n roots in the complex domain. However, there may be fewer roots in the real domain. This is because if the coefficients are real then there may be pairs of complex roots [conjugates] which will not count as real roots. Also, there may be identical roots of multiple order. For example, x4 - 1 = 0 has 4 complex roots. These are 1, -1, i and -i where i is the imaginary root of -1. There are only 2 real roots -1 and +1. x4 = 0 has 4 multiple roots, each of which is 0. Thus x = 0 is a root of multiplicity 4.
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
If you meant x + 5 = 4 then the multiplicity is 1.
A quadratic polynomial.
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 ...
4
4, the same as the degree of the polynomial.
Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)
Each distinct real root is an x-intercept. So the answer is 4.
There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.
Upto 4. If the coefficients are all real, then it can have only 0, 2 or 4 real roots.
None, it involves the square root of a negative number so the roots are imaginary.
(x - u)*(x - u)*(x + 2i)*(x - 2i) = (x2 - 2xu + u2)*(x2 + 4) = x4 - 2x3u + x2(u2 + 4) - 8xu + 4u2
A rational equation can be multiplied by the least common multiple of its denominators to make it into a polynomial equation. The degree of this polynomial is the highest power (of the variable) that appears in it. It can be proven that a polynomial of degree n must have n roots in the complex domain. However, there may be fewer roots in the real domain. This is because if the coefficients are real then there may be pairs of complex roots [conjugates] which will not count as real roots. Also, there may be identical roots of multiple order. For example, x4 - 1 = 0 has 4 complex roots. These are 1, -1, i and -i where i is the imaginary root of -1. There are only 2 real roots -1 and +1. x4 = 0 has 4 multiple roots, each of which is 0. Thus x = 0 is a root of multiplicity 4.
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
It is difficult to tell because there is no sign (+ or -) before the 5. +5 gives complex roots and assuming that someone who asked this question has not yet come across complex numbers, I assume the polynomial is x2 -3x - 5 The roots of this equation are: -1.1926 and 4.1926 (to 4 dp)