Find the smaller positive integer and the largest negative integer that, by the

Question

Find the smaller positive integer and the largest negative integer that, by the Upper and tower Bound Theorem, are upper and lower hounds (or the real zeros of the polynomial function F(x) = 2x^3 + x^2 - 24 x + 15 Use Descartes' Rule o(Signs to state the number of possible positive and negative real zeros of the polynomial function. (Enter your answers as a comma separated list.) P(x) = x^5 + x^4 + 4x^3 - 4^2 + 18x + 18 number of possible positive real zeros number of possible negative zeal zeros

Explanation / Answer

As per the upper and lower bounds theorem:

Also, as per the rational roots theorem, if P(x) is a polynomial with integer coefficients and if p/q is a zero of P(x) , then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .

Here , P(x) = 2x3 + x2 -24x + 15. Thus, as per the rational roots theorem, the zeros of P(x) are p/q , where p can be ± 1, ±3, ±5 or ±15 and q can be ±1 or ±2. Thus, the lowest and the highest values for p/q are – 15 and + 15 respectively. We will now, divide P(x) by (x +15) and (x - 15), using synthetic division, to ascertain the lower and the upper bounds for P(x). On dividing P(x) by ( x- 15),using synthetic division, we get +2x2, + 31x and +441 on the bottom row and the remainder is 6630. Since there is no negative entry in the bottom row, 15 is an upper bound for P(x). Now, if we divide P(x) by [ x – (-15)] i.e. (x + 15), using synthetic division, we get +2x2 , - 14 and + 186 on the bottom row and the remainder is – 2775. Thus, -15 is a lower bound for P(x).

As per Descartes' rule of signs, if the terms of a single-variable polynomial with real coefficients are arranged in descending order ( exponent wise), then the number of positive roots of the polynomial is equal to the number of changes of sign between the consecutive nonzero coefficients, or is less than this number by an even number.

Here, P(x) = x5 + x4 + 4x3 – 4x2 + 18x + 18. We can observe that the coefficients are +1, +1, +4, -4 , +18 and +18 so there are 2 changes of sign. Thus, the maximum number of possible positive real zeros of P(x) is 2. Further P (-x) = -x5 + x4 – 4x3 -4x2 – 18x +18. We can observe that there are 3 changes of sign here. Thus, the maximum number of possible negative real zeros of P(x) is 3.

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