,MAT1511 Assignment 2 Semester 1 , Due April 2025,,.100%
TRUSTED Complete, trusted solutions and explanations.
WE WISH YOU ALL THE BEST
QUESTION 1 1.1 Use Descartes’ Rule of Signs to describe all
possibilities for the number of positive, negative and imaginary
zeros of P (x) = x4 + x3 + x2 + x + 12 (Summarize your answer
in the form of a table like the example on p. 297). (4)
1.2 P (x) = x4 − 2x3 − 2x2 − 2x − 3
(a) Use the Upper and Lower Bounds Theorem to show that all
zero of P (x) are bounded below by −1 and above by 3.
(b) Find all the possible rational zeros of P (x) by using the
Rational Zero Theorem. (1) (c) Solve P (x) = 0 (i.e. find all the
solutions of P (x) = 0.) (3) [11]
𝟏. 𝟏 𝑫𝒆𝒔𝒄𝒂𝒓𝒕𝒆𝒔′ 𝑹𝒖𝒍𝒆 𝒐𝒇 𝑺𝒊𝒈𝒏𝒔 𝒇𝒐𝒓 𝒕𝒉𝒆 𝑷𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐:
Descartes' Rule of Signs:
This rule provides a way to estimate the number of positive and
negative real zeros of a polynomial. Here's how it works:
Positive zeros: The number of positive real zeros is either
equal to the number of sign changes between consecutive
non-zero terms of the polynomial, or less than that by an
even number.
, Negative zeros: The number of negative real zeros is equal
to the number of sign changes in
𝑷(−𝒙)𝑷(−𝒙)𝑃(−𝑥), 𝑜𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑡ℎ𝑎𝑡 𝑏𝑦 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟.
𝑺𝒕𝒆𝒑 𝟏: 𝑷𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒁𝒆𝒓𝒐𝒔 𝒐𝒇 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐
𝑊𝑒 𝑒𝑥𝑎𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑜𝑓 𝑷(𝒙)𝑷(𝒙)𝑃(𝑥):
𝑇ℎ𝑒 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑖𝑠:
𝑷(𝒙) = 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑃(𝑥)
= 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 12
𝑇ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑎𝑟𝑒: + 𝟏, +𝟏, +𝟏, +𝟏, +𝟏𝟐 +
𝟏, +𝟏, +𝟏, +𝟏, +𝟏𝟐 + 1, +1, +1, +1, +12
𝑺𝒕𝒆𝒑 𝟐: 𝑵𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒁𝒆𝒓𝒐𝒔 𝒐𝒇 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐
𝑁𝑜𝑤, 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑷(−𝒙)𝑷(−𝒙)𝑃(−𝑥):
𝑷(−𝒙) = (−𝒙)𝟒 + (−𝒙)𝟑 + (−𝒙)𝟐 + (−𝒙) +
𝟏𝟐𝑷(−𝒙) = (−𝒙)^𝟒 + (−𝒙)^𝟑 + (−𝒙)^𝟐 + (−𝒙) +
𝟏𝟐𝑃(−𝑥) = (−𝑥)4 + (−𝑥)3 + (−𝑥)2 + (−𝑥) + 12
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔:
𝑷(−𝒙) = 𝒙𝟒 − 𝒙𝟑 + 𝒙𝟐 − 𝒙 + 𝟏𝟐𝑷(−𝒙)
= 𝒙^𝟒 − 𝒙^𝟑 + 𝒙^𝟐 − 𝒙 + 𝟏𝟐𝑃(−𝑥)
= 𝑥4 − 𝑥3 + 𝑥2 − 𝑥 + 12
TRUSTED Complete, trusted solutions and explanations.
WE WISH YOU ALL THE BEST
QUESTION 1 1.1 Use Descartes’ Rule of Signs to describe all
possibilities for the number of positive, negative and imaginary
zeros of P (x) = x4 + x3 + x2 + x + 12 (Summarize your answer
in the form of a table like the example on p. 297). (4)
1.2 P (x) = x4 − 2x3 − 2x2 − 2x − 3
(a) Use the Upper and Lower Bounds Theorem to show that all
zero of P (x) are bounded below by −1 and above by 3.
(b) Find all the possible rational zeros of P (x) by using the
Rational Zero Theorem. (1) (c) Solve P (x) = 0 (i.e. find all the
solutions of P (x) = 0.) (3) [11]
𝟏. 𝟏 𝑫𝒆𝒔𝒄𝒂𝒓𝒕𝒆𝒔′ 𝑹𝒖𝒍𝒆 𝒐𝒇 𝑺𝒊𝒈𝒏𝒔 𝒇𝒐𝒓 𝒕𝒉𝒆 𝑷𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐:
Descartes' Rule of Signs:
This rule provides a way to estimate the number of positive and
negative real zeros of a polynomial. Here's how it works:
Positive zeros: The number of positive real zeros is either
equal to the number of sign changes between consecutive
non-zero terms of the polynomial, or less than that by an
even number.
, Negative zeros: The number of negative real zeros is equal
to the number of sign changes in
𝑷(−𝒙)𝑷(−𝒙)𝑃(−𝑥), 𝑜𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑡ℎ𝑎𝑡 𝑏𝑦 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟.
𝑺𝒕𝒆𝒑 𝟏: 𝑷𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒁𝒆𝒓𝒐𝒔 𝒐𝒇 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐
𝑊𝑒 𝑒𝑥𝑎𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑜𝑓 𝑷(𝒙)𝑷(𝒙)𝑃(𝑥):
𝑇ℎ𝑒 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑖𝑠:
𝑷(𝒙) = 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑃(𝑥)
= 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 12
𝑇ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑎𝑟𝑒: + 𝟏, +𝟏, +𝟏, +𝟏, +𝟏𝟐 +
𝟏, +𝟏, +𝟏, +𝟏, +𝟏𝟐 + 1, +1, +1, +1, +12
𝑺𝒕𝒆𝒑 𝟐: 𝑵𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒁𝒆𝒓𝒐𝒔 𝒐𝒇 𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙^𝟒 + 𝒙^𝟑 + 𝒙^𝟐 + 𝒙 + 𝟏𝟐𝑷(𝒙)
= 𝒙𝟒 + 𝒙𝟑 + 𝒙𝟐 + 𝒙 + 𝟏𝟐
𝑁𝑜𝑤, 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑷(−𝒙)𝑷(−𝒙)𝑃(−𝑥):
𝑷(−𝒙) = (−𝒙)𝟒 + (−𝒙)𝟑 + (−𝒙)𝟐 + (−𝒙) +
𝟏𝟐𝑷(−𝒙) = (−𝒙)^𝟒 + (−𝒙)^𝟑 + (−𝒙)^𝟐 + (−𝒙) +
𝟏𝟐𝑃(−𝑥) = (−𝑥)4 + (−𝑥)3 + (−𝑥)2 + (−𝑥) + 12
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔:
𝑷(−𝒙) = 𝒙𝟒 − 𝒙𝟑 + 𝒙𝟐 − 𝒙 + 𝟏𝟐𝑷(−𝒙)
= 𝒙^𝟒 − 𝒙^𝟑 + 𝒙^𝟐 − 𝒙 + 𝟏𝟐𝑃(−𝑥)
= 𝑥4 − 𝑥3 + 𝑥2 − 𝑥 + 12
