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(3x - 1)(x - 4) so roots are 1/3 and 4
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What is the polynomial roots to 1 5i and -3i?
4
How can a rational equation have more than one solution?
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.
What is the discriminant in the polynomial x2 plus 4x plus 5?
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
What is the multiplicity of the zero of the polynomial function f x x plus 5 4?
If you meant x + 5 = 4 then the multiplicity is 1.
What type of polynomial is 2x2-6x 4?
A quadratic polynomial.
what are all of the zeros of this polynomial function f(a)=a^4-81?
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 ...
What is the polynomial roots to 1 5i and -3i?
4
Is it true that the degree of polynomial function determine the number of real roots?
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.)
At most how many unique roots will a fourth degree polynomial have?
4, the same as the degree of the polynomial.
How many x-intercepts does a quartic polynomial function having 4 distinct real roots have?
Each distinct real root is an x-intercept. So the answer is 4.
Which polynomial has rational coefficients a leading leading coefficient of 1 and the zeros at 2-3i and 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.
How many real roots can a fourth degree polynomial have?
Upto 4. If the coefficients are all real, then it can have only 0, 2 or 4 real roots.
The polynomial 4x2 plus 5x plus 4 has how many roots?
None, it involves the square root of a negative number so the roots are imaginary.
What is the quartic polynomial function with rational coefficients that has roots you and 2i?
(x - u)*(x - u)*(x + 2i)*(x - 2i) = (x2 - 2xu + u2)*(x2 + 4) = x4 - 2x3u + x2(u2 + 4) - 8xu + 4u2
How can a rational equation have more than one solution?
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.
What is the discriminant in the polynomial x2 plus 4x plus 5?
It is: 16-20 = -4 which means that the given quadratic expression has no real roots.
Which two values of x are roots of the polynomial below x2-3x 5?
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)
