Question 1209771: Suppose the polynomial p(x)=x^3+ax^2+bc+c has the property that the mean of its zeroes, the product of its zeroes, and the sum of its coefficients are all equal. If the y-intercept of the graph of y=p(x) is 0, what is b?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step:
**1. Define the Zeros:**
* Let the zeros of p(x) be r, s, and t.
**2. Relationships Between Zeros and Coefficients:**
* **Mean of Zeros:** (r + s + t) / 3 = -a / 3
* **Product of Zeros:** rst = -c
* **Sum of Coefficients:** p(1) = 1 + a + b + c
**3. Given Conditions:**
* The mean of the zeros, the product of the zeros, and the sum of the coefficients are all equal.
* The y-intercept of p(x) is 0. This means p(0) = 0.
**4. Use the Y-Intercept Condition:**
* p(0) = 0^3 + a(0)^2 + b(0) + c = 0
* c = 0
**5. Apply the Equal Condition:**
* Since c = 0, the product of the zeros is rst = -c = 0. This implies at least one of the zeros is 0. Let's say r = 0.
* Now, we have:
* (r + s + t) / 3 = -a / 3
* rst = 0
* 1 + a + b + c = 1 + a + b
* Since all three are equal:
* -a / 3 = 0
* 1 + a + b = 0
**6. Solve for a and b:**
* From -a / 3 = 0, we get a = 0.
* Substitute a = 0 into 1 + a + b = 0:
* 1 + 0 + b = 0
* b = -1
**7. Check the Mean of Zeros:**
* Since a = 0, the mean of the zeros is -a / 3 = 0.
* Since r = 0, we have (0 + s + t) / 3 = 0, which means s + t = 0.
* Also, 1 + a + b = 1 + 0 - 1 = 0, which confirms the condition.
**Answer:**
The value of b is -1.
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