Find the smallest value for the polynomial

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The discussion revolves around finding the smallest value among various expressions related to the quartic polynomial P(x) = x^4 + ax^3 + bx^2 + cx + d. Key calculations include P(-1), the product of the zeros (d), and the sum of the coefficients (1 + a + b + c + d). Participants analyze the positivity of the product of real zeros and non-real zeros, concluding that P(-1) and the product of zeros are positive, while the sum of coefficients could potentially be negative. By evaluating the graph and derivatives, they suggest that the product of the non-real zeros is likely the smallest value, estimating it to be less than 2. The analysis emphasizes the importance of eliminating impossible options based on the polynomial's characteristics.
Loststudent22
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The graph below shows a portion of the curve defined by the quartic polynomial P(x) = x^4 + ax^3 + bx^2 + cx + d. Which of the following is the smallest?https://imgur.com/a/1VuGSiA(A) P(-1) (B) The product of the zeros of P (C) The product of the non-real zeros of P (D) The sum of the coefficients of P (E) The sum of the real zeros of PI know that P(-1) = 1-a+b-c+d Product of zeroes is d. Real zeroes are around 1.7 and 3.85, so product of non-reals is d/(1.7*3.85) Sum of the coefficients is 1+a+b+c+d. Sum of the zeros is -a and that P(0)=d and P(1)=1+a+b+c+d. How am I supposed to tell which is smallest with this information though?
 
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Loststudent22 said:
The graph below shows a portion of the curve defined by the quartic polynomial P(x) = x^4 + ax^3 + bx^2 + cx + d. Which of the following is the smallest?https://imgur.com/a/1VuGSiA(A) P(-1) (B) The product of the zeros of P (C) The product of the non-real zeros of P (D) The sum of the coefficients of P (E) The sum of the real zeros of PI know that P(-1) = 1-a+b-c+d Product of zeroes is d. Real zeroes are around 1.7 and 3.85, so product of non-reals is d/(1.7*3.85) Sum of the coefficients is 1+a+b+c+d. Sum of the zeros is -a and that P(0)=d and P(1)=1+a+b+c+d. How am I supposed to tell which is smallest with this information though?
I think that this problem requires you to elliminate the potential answers that could not be answers. For example, both real zeroes are positive, so their product would also be positive. Any nonreal zeroes have to occur in conjugate pairs. For this problem I believe that the nonreal zeroes would be purely imaginary; if so, their product would also be positive. Can you eliminate any more possibilities in continuing with this kind of analysis?

Also, even though this was posted in the Precalc section, I suspect that more information can be obtained by looking at the first and second derivatives, noticing where both the first and second derivatives are negative, positive, or zero.

What's the context for this problem? Is it in a textbook of some kind? Was the problem given in a math class? If so, was it a calculus class?
 
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Mark44 said:
I think that this problem requires you to elliminate the potential answers that could not be answers. For example, both real zeroes are positive, so their product would also be positive. Any nonreal zeroes have to occur in conjugate pairs. For this problem I believe that the nonreal zeroes would be purely imaginary; if so, their product would also be positive. Can you eliminate any more possibilities in continuing with this kind of analysis?
Well from the graph P(-1) is positive so I can eliminate that. Also at P(0) I know that D is also positive. Yet at P(1) the graph get smaller so I can assume that a and c are probably negative which would mean that the sum of the cofficients might be negative?It was a high school contest problem I came across so I assume that calculus could be used.
 
Hmm c is the product of the x-values of the minima and maxima of the function (zeroes of the derivative). c is either 0 or slightly negative. So would that mean its the correct choice then?
 
The sum of the real zeroes is at least 4, while the product of the non-real zeroes is d divided by the product of the real zeroes, which is smaller than 6/3 = 2, which beats out all the other ones.
 
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