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

Solution:

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

Solution:

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

Solution:

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)