MAT1511 Assignment 2 (COMPLETE ANSWERS) 2025 - DUE April 2025; 100% TRUSTED Complete, trusted solutions and explanations.

,MAT1511 Assignment 2 (COMPLETE ANSWERS) 2025 -
DUE April 2025; 100% TRUSTED Complete, trusted
solutions and explanations. ……WE WISH YOU A GOOD
LUCK
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.


(a) Determine whether (x−2)(x - 2)(x−2) is a factor of
P(x)P(x)P(x)
𝑇𝑜 𝑐ℎ𝑒𝑐𝑘 𝑤ℎ𝑒𝑡ℎ𝑒𝑟 (𝑥 − 2)(𝑥 − 2)(𝑥
− 2) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥), 𝑤𝑒 𝑐𝑎𝑛 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑭𝒂𝒄𝒕𝒐𝒓 𝑻𝒉𝒆𝒐𝒓𝒆𝒎
− 𝑐)(𝑥 − 𝑐)(𝑥
− 𝑐) 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑃(𝑐)
= 0𝑃(𝑐) = 0𝑃(𝑐) = 0.
𝑆𝑜, 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑃(2)𝑃(2)𝑃(2) 𝑎𝑛𝑑 𝑐ℎ𝑒𝑐𝑘 𝑖𝑓 𝑖𝑡 𝑒𝑞𝑢𝑎𝑙𝑠 0.
𝑇ℎ𝑒 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦:
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
𝐿𝑒𝑡′𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = 2𝑥 = 2𝑥 = 2 𝑖𝑛𝑛𝑜 𝑡ℎ𝑒 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙:

, 𝑃(2) = (2)6 − 2(2)5 − (2)4 + (2)3 + 2(2)2 + 2 − 2𝑃(2)
= (2)^6 − 2(2)^5 − (2)^4 + (2)^3 + 2(2)^2
+ 2 − 2𝑃(2)
= (2)6 − 2(2)5 − (2)4 + (2)3 + 2(2)2 + 2 − 2
= 64 − 2(32) − 16 + 8 + 2(4) + 2 − 2
= 64 − 2(32) − 16 + 8 + 2(4) + 2 − 2
= 64 − 2(32) − 16 + 8 + 2(4) + 2 − 2
= 64 − 64 − 16 + 8 + 8 + 2 − 2
= 64 − 64 − 16 + 8 + 8 + 2 − 2
= 64 − 64 − 16 + 8 + 8 + 2 − 2
= 64 − 64 − 16 + 8 + 8 + 2 − 2 = 0
= 64 − 64 − 16 + 8 + 8 + 2 − 2 = 0
= 64 − 64 − 16 + 8 + 8 + 2 − 2 = 0
𝑆𝑖𝑛𝑐𝑒 𝑃(2) = 0𝑃(2) = 0𝑃(2)
= 0, (𝑥 − 2)(𝑥 − 2)(𝑥
− 2) 𝒊𝒔 𝒊𝒏𝒅𝒆𝒆𝒅 𝒂 𝒇𝒂𝒄𝒕𝒐𝒓 𝑜𝑓 𝑃(𝑥)𝑃(𝑥)𝑃(𝑥).


(b) Find all the possible rational zeros of P(x)P(x)P(x) using
the Rational Zeros Theorem
The Rational Zeros Theorem states that if a polynomial has a
rational root pq\frac{p}{q}qp, then:
 ppp is a factor of the constant term (the term without xxx,
in this case, −2-2−2).
 qqq is a factor of the leading coefficient (the coefficient of
x6x^6x6, in this case, 111).
Step 1: Find the factors of the constant term (−2-2−2)