Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3+x2–2x–1, g(x) = x + 1
(ii) p(x) = x3+3x2+3x+1, g(x) = x + 2
(iii) p(x) = x3–4x2+x+6, g(x) = x – 3
(i) p(x) = 2x3+x2–2x–1, g(x) = x+1
p(x) = 2x3+x2–2x–1, g(x) = x+1
g(x) = 0
⇒ x + 1 = 0
⇒ x = −1
∴Zero of g(x) is -1.
Now,
p(−1) = 2(−1)3+(−1)2–2(−1)–1
= −2+1+2−1
= 0
∴By factor theorem, g(x) is a factor of p(x).
(ii) p(x) = x3+3x2+3x+1, g(x) = x+2
p(x) = x3+3x2+3x+1, g(x) = x+2
g(x) = 0
⇒ x+2 = 0
⇒ x = −2
∴ Zero of g(x) is -2.
Now,
p(−2) = (−2)3+3(−2)2+3(−2)+1
= −8+12−6+1
= −1 ≠ 0
∴By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x3–4x2+x+6, g(x) = x–3
p(x) = x3–4x2+x+6, g(x) = x -3
g(x) = 0
⇒ x−3 = 0
⇒ x = 3
∴ Zero of g(x) is 3.
Now,
p(3) = (3)3−4(3)2+(3)+6
= 27 − 36 + 3 + 6
= 0
∴By factor theorem, g(x) is a factor of p(x).