3x^3 -15x^2 -13x-4
what are the zeros of this problem?
when x=6 the expression is positive,
when x=5 it is negative, so there is a zero between 5 and 6
I ran it trough my homemade equation solver and got x = appr. 5.788
reducing the expression using synthetic division resulted in two imaginary roots.
So the expression has one real zero at about x = 5.788
The "p/q theorem" says that if there is any real rational root, it will be + or - p/q, where p and q are prime factors of the first and last terms, which are 3 and 1 for p and 1,2 and 4 for q. I can't find any p/q ratios that work, so a graphical or iterative (trial and error) solution may be required.
To find the zeros of the polynomial 3x^3 - 15x^2 - 13x - 4, we need to solve for x when the polynomial is equal to zero.
One way to find the zeros is by using the Rational Root Theorem and synthetic division. This theorem states that any rational zero of the polynomial must be a factor of the constant term (in this case, -4) divided by a factor of the leading coefficient (in this case, 3).
The factors of -4 are ±1, ±2, and ±4, and the factors of 3 are ±1 and ±3. By performing synthetic division with these possible rational zeros, we can determine if they are actually zeros or not.
Let's begin with the possible rational zero x = 1.
1 | 3 -15 -13 -4
| 3 -12 -1
|_________________
3 -12 -25 -5
Since the remainder is not zero, x = 1 is not a zero of the polynomial.
Next, let's try x = -1.
-1 | 3 -15 -13 -4
| -3 18 -5
|_________________
3 -18 5 -9
Again, the remainder is not zero, so x = -1 is not a zero either.
We can continue this process by trying x = 2, -2, 4, and -4, or we can use a graphing calculator to find the zeros. By graphing the polynomial, we can visually identify the x-intercepts or zeros.
Alternatively, we can use a numerical method like Newton's method or the bisection method to estimate the zeros, or we can use factoring techniques if the polynomial is factorable.
Based on the calculations or graphing, we can determine the zeros of the polynomial.