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Mathematics

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

Polynomials

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Answer

(i) p(x) = 2x3 + x2 - 2x - 1

g(x) = x + 1

⇒ x + 1 = 0

⇒ x = -1

Putting x = -1 we get,

p(-1) = 2(-1)3 + (-1)2 - 2(-1) - 1

= -2 + 1 + 2 -1

= 0

Remainder is zero (0), so g(x) is factor of p(x).

(ii) p(x) = x3 + 3x2 + 3x + 1

g(x) = x + 2

⇒ x + 2 = 0

⇒ x = -2

Putting x = -2 we get,

p(-2) = (-2)3 + 3(-2)2 + 3(-2) + 1

= -8 + 3(4) - 6 + 1

= -8 + 12 - 6 + 1

= 13 - 14

= -1

Remainder is not zero (0), so g(x) is not a factor of p(x).

(iii) p(x) = x3 - 4x2 + x + 6

g(x) = x - 3

⇒ x - 3 = 0

⇒ x = 3

Putting x = -3 we get,

p(3) = (3)3 - 4(3)2 + 3 + 6

= 27 - 4(9) + 9

= 27 - 36 + 9

= 36 - 36

= 0

Remainder is zero (0), so g(x) is a factor of p(x).

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