Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero( s):
-1, radical 3, 11/3
One of the rational roots would have to be a double root, so it could be
y = (x+1)^2(x-√3)(x+√3)(3x-11) = (x+1)^2(x^2+9)(3x-11)
or
y = (x+1)(x-√3)(x+√3)(3x-11)^2 = (x+1)(x^2+9)(3x-11)^2
Suppose that a polynomial function of degree 5 with rational coefficients has -2, 5, and 3 - I as zeros. Find the other zeros.
(x+1)(x-√3)(x+√3)(3x-11)
no way to tell the 5th root.
did I miss an i somewhere?
(x-√3)(x+√3) = x^2-3
To find the other zeros of a polynomial function, we can use the fact that the zeros of a polynomial are the values of x that make the polynomial equal to zero.
Let's denote the polynomial as P(x) of degree 5 with rational coefficients. We are given three zeros: -1, √3, and 11/3.
Since the polynomial has rational coefficients, the conjugates of the irrational zeros will also be zeros of the polynomial.
Therefore, the other zero will be the conjugate of √3, which is -√3.
Now we have four zeros: -1, √3, 11/3, and -√3.
To find the last zero, we know that the sum of the zeros of a polynomial of degree 5 is equal to the opposite of the coefficient of the x^4 term divided by the coefficient of the x^5 term.
Let's represent the polynomial as:
P(x) = a * (x + 1) * (x - √3) * (x + √3) * (x - 11/3) * (x - z)
where 'z' represents the unknown zero.
Since P(x) is of degree 5, the coefficient of the x^5 term is 'a' (non-zero).
We know that the sum of the zeros of P(x) is equal to -(-1/a) = 1/a.
Therefore, 1/a = -1 + √3 + (-√3) + 11/3 + z
Simplifying this equation, we get:
1/a = 11/3 + z
We can rearrange this equation to find 'z':
z = 1/a - 11/3
Now, we have the value of 'z' as 1/a - 11/3.
Hence, the other zero of the polynomial function is 1/a - 11/3, where 'a' is the coefficient of the x^5 term.