MAT1511 Assignment 2 (COMPLETE ANSWERS) 2025 - DUE April 2025; 100% correct solutions and explanations.

, MAT1511 Assignment 2 (COMPLETE ANSWERS) 2025 - DUE April 2025; 100% correct
solutions and explanations.
1. Let P(x) = x6−2x5−x4+x3+2x2+x−2
(a) Determine whether (x−2)is a factor of P(x). (2)
(b) Find all the possible rational zeros of P(x)by using the Rational Zeros Theorem. (2)
(c) Solve P(x) = 0. (4)3.



𝐿𝑒𝑡′𝑠 𝑤𝑜𝑟𝑘 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑒𝑎𝑐ℎ 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 𝑠𝑡𝑒𝑝 𝑏𝑦 𝑠𝑡𝑒𝑝:

𝑇𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑤ℎ𝑒𝑡ℎ𝑒𝑟 (𝑥 − 2)(𝑥 − 2)(𝑥
− 2) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥), 𝑤𝑒 𝑐𝑎𝑛 𝑢𝑠𝑒 𝒕𝒉𝒆 𝑭𝒂𝒄𝒕𝒐𝒓 𝑻𝒉𝒆𝒐𝒓𝒆𝒎. 𝑇ℎ𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑠𝑡𝑎𝑡𝑒𝑠 𝑡ℎ𝑎
− 𝑐)(𝑥 − 𝑐)(𝑥 − 𝑐) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑎 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑃(𝑐) = 0𝑃(𝑐) = 0𝑃(𝑐)
= 0.

(𝒂) 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝒘𝒉𝒆𝒕𝒉𝒆𝒓 (𝒙 − 𝟐)(𝒙 − 𝟐)(𝒙 − 𝟐) 𝒊𝒔 𝒂 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝑷(𝒙)𝑷(𝒙)𝑷(𝒙).

𝑇ℎ𝑢𝑠, 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑃(2)𝑃(2)𝑃(2):

𝑃(𝑥) = 𝑥6 − 2𝑥5 − 𝑥4 + 𝑥3 + 2𝑥2 + 𝑥 − 2𝑃(𝑥)
= 𝑥^6 − 2𝑥^5 − 𝑥^4 + 𝑥^3 + 2𝑥^2 + 𝑥 − 2𝑃(𝑥)
= 𝑥6 − 2𝑥5 − 𝑥4 + 𝑥3 + 2𝑥2 + 𝑥 − 2

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 2𝑥 = 2𝑥 = 2 𝑖𝑛𝑡𝑜 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥):

𝑃(2) = 26 − 2 ⋅ 25 − 24 + 23 + 2 ⋅ 22 + 2 − 2𝑃(2)
= 2^6 − 2 \𝑐𝑑𝑜𝑡 2^5 − 2^4 + 2^3 + 2 \𝑐𝑑𝑜𝑡 2^2 + 2 − 2𝑃(2)
= 26 − 2 ⋅ 25 − 24 + 23 + 2 ⋅ 22 + 2 − 2 𝑃(2)
= 64 − 64 − 16 + 8 + 8 + 2 − 2𝑃(2)
= 64 − 64 − 16 + 8 + 8 + 2 − 2𝑃(2)
= 64 − 64 − 16 + 8 + 8 + 2 − 2 𝑃(2) = 64 − 64 − 16 + 8 + 8 + 2 − 2
= 0𝑃(2) = 64 − 64 − 16 + 8 + 8 + 2 − 2 = 0𝑃(2)
= 64 − 64 − 16 + 8 + 8 + 2 − 2 = 0

𝑆𝑖𝑛𝑐𝑒 𝑃(2) = 0𝑃(2) = 0𝑃(2)
= 0, 𝑏𝑦 𝑡ℎ𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 𝑇ℎ𝑒𝑜𝑟𝑒𝑚, (𝑥 − 2)(𝑥 − 2)(𝑥
− 2) 𝒊𝒔 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥).

(𝒃) 𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒕𝒉𝒆 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒛𝒆𝒓𝒐𝒔 𝒐𝒇 𝑷(𝒙)𝑷(𝒙)𝑷(𝒙) 𝒃𝒚 𝒖𝒔𝒊𝒏𝒈 𝒕𝒉𝒆 𝑹𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒁𝒆𝒓𝒐𝒔 𝑻𝒉𝒆𝒐𝒓