how to find all real zeros of "4x^5+8x^4-15x^3-23x^2+11x+15
To find all real zeros of a polynomial, such as the one you provided (4x^5 + 8x^4 - 15x^3 - 23x^2 + 11x + 15), you can use a combination of the rational root theorem and synthetic division. Here's a step-by-step guide:
Step 1: Identify the coefficients of the polynomial:
The given polynomial is: 4x^5 + 8x^4 - 15x^3 - 23x^2 + 11x + 15
Step 2: Apply the rational root theorem:
The rational root theorem suggests that if there are any rational roots (zeros) of the polynomial, they will be in the form of p/q, where p is a factor of the constant term (15 in this case), and q is a factor of the leading coefficient (4 in this case).
List out all the possible values for p and q. For p, the factors of 15 are ±1, ±3, ±5, ±15. For q, the factors of 4 are ±1 and ±2.
The possible rational roots of the polynomial are: ±1/1, ±3/1, ±5/1, ±15/1, ±1/2, ±3/2, ±5/2, ±15/2.
Step 3: Use synthetic division:
To test each possible rational root, you can use synthetic division. Synthetic division helps you quickly evaluate whether a particular value is actually a root (zero) of the polynomial.
For example, let's test the value -1. Begin by writing down the coefficients of the polynomial:
4 | 4 8 -15 -23 11 15
Using synthetic division, divide each coefficient by 4 and perform the operation as follows:
-1 | 4 8 -15 -23 11 15
-------------------------------------
| -4 -4 19 -4 -7
-------------------------------------
4 4 -19 -4 7 8
The final row of the synthetic division represents the coefficients of the resulting polynomial after dividing by (x + 1). In this case, the coefficients are 4, 4, -19, -4, 7, 8. Since the last coefficient is not zero, -1 is not a root of the polynomial.
Repeat this process for the other possible rational roots you listed.
By using the rational root theorem and synthetic division, you can test all possible rational roots and find the ones that produce a remainder of 0, indicating that they are zeros of the polynomial.
Since there is no general method to solve a quintic equation, we must assume that there is some low-hanging fruit, root-wise.
y(1) = 0, so
y = (x-1)(4x^4 + 12x^3 - 3x^2 - 26x - 15)
Try 1 again: no joy
Try -1. Yay and we have
y = (x-1)(x+1)(4x^3 + 8x^2 - 11x - 15)
Try -1 again: yay
y = (x-1)(x+1)(x+1)(4x^2 + 4x - 15)
y = (x-1)(x+1)(x+1)(2x-3)(2x+5)
I assume you can find the roots at this point...