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Chapter 2: Problem 29

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=2 x^{3}+x^{2}-25 x+12$$

Short Answer

Expert verified
The number of possible positive real zeros of the polynomial P(x)=2x^3+x^2-25x+12 is either 2 or 0. The number of possible negative real zeros is either 1 or 0.

Step by step solution

01

Identify the coefficients and their signs

The polynomial \(P(x)=2 x^{3}+x^{2}-25 x+12\) has coefficients 2, 1, -25 and 12. Their signs are +, +, -, + respectively.
02

Apply Descartes' Rule of Signs for positive real zeros

The number of sign changes in the polynomial will give the number of positive real zeros. Here, we observe two changes in signs: from + to - and - to +. So the number of positive real zeros in the polynomial is 2 or 0 (2 subtracted by a multiple of 2, and it can't be negative).
03

Apply Descartes' Rule of Signs for negative real zeros

Now replace 'x' with '-x' in the polynomial to find the number of negative real zeros. The polynomial becomes \( P(-x)=2(-x)^3 + (-x)^2 -25 (-x) + 12\), which simplifies to: -2x^3 + x^2 + 25x + 12. The coefficients of this new polynomial are now -, +, +, +. There is only one change in sign, from - to +. Therefore, the number of negative real zeros is either 1 or 0.

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