MAT1511 Assignment 1 (ANSWERS) Semester 1 - 2023.
Questions asked: 1. Use Descartes’ Rule of Signs to determine the possible number of positive, negative and
imaginary zeros of P (x).
(i) P (x) = x
4 +x
3 +x
2 +x +12. (4)
(ii) P (x) = 2x
5 +x
4 +x
3 −4x
2 −x −3. (4)
2. Let P ...
,1 ADDENDUM A: YEAR ASSIGNMENT 01 3.2 – 3.5, 8.1, 8.3, 8.4 FIXED CLOSING DATE: 18 May 2023
UNIQUE NUMBER: 768575 NO EXTENSION CAN BE GRANTED 1. Use Descartes’ Rule of Signs to
determine the possible number of positive, negative and imaginary zeros of P (x). (i) P (x) = x 4 +x 3 +x 2
+x +12. (4) (ii) P (x) = 2x 5 +x 4 +x 3 −4x 2 −x −3. (4)
(i) To use Descartes' Rule of Signs to determine the possible number of positive, negative, and
imaginary zeros of P(x) = x^4 + x^3 + x^2 + x + 12, we need to count the sign changes in the coefficients
of the polynomial.
According to Descartes' Rule of Signs, the number of positive zeros of P(x) is either equal to the number of sign
changes or less than it by an even number. In this case, there are 1 sign change in the coefficients, so P(x) has
either 1 positive zero or 3 positive zeros (less by an even number).
To determine the number of negative zeros, we substitute -x for x in P(x) and count the sign changes:
According to Descartes' Rule of Signs, the number of negative zeros of P(x) is either equal to the number of
sign changes or less than it by an even number. In this case, there are 3 sign changes in the coefficients, so
P(x) has either 3 negative zeros or 1 negative zero (less by an even number).
Since the number of positive zeros and negative zeros can differ by an even number, the possible combinations
for the number of positive, negative, and imaginary zeros are:
- 1 positive zero and 3 negative zeros (or vice versa).
- 3 positive zeros and 1 negative zero (or vice versa).
- 1 positive zero, 1 negative zero, and 2 imaginary zeros.
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