Question

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

Answer 1

Answer:

(i) p(x) = 2x3 + x2 – 2x – 1 g(x) = x + 1 If x + 1 = 0, then x = -1 p(x) = 2x3 + x2 – 2x – 1 p(-1)= 2(-1)3 + (-1)2 -2(- 1) – 1 = 2(-1) + (1) + 2 – 1 = -2 + 1 + 2 – 1 P(-1)= 0 Here, r(x) = p(a) = 0, ∴ g(x) is the a factor f(x)

(ii) p(x) = x3 + 3x2 + 3x + 1 g(x) = x + 2 If x + 2 = 0, then x = -2 ∴ p(x) = x3 + 3x2 + 3x + 1 p(-2) = (-2)3 + 3(-2)2 + 3 (-2) + 1 = -8 + 3(4) + 3(-2) + 1 = -8 + 12 – 6 + 1 = 13 – 24 p(-2)= -11 Here we have r(x) = p(a) =-11. Value of r(x) is not equal to zero. ∴ g(x) is not a factor of f(x).

(iii) p(x) = x3 – 4x2 + x + 6 g(x) = x – 3 Let, x – 3 = 0, then x = 3 p(x) = x3 – 4x2 + x + 6 p(3) = (3)2 – 4(3)2 + 3 + 6 = 27 – 4(9) + 3 + 6 = 27 – 36 + 3 + 6 = 36 – 36 p(3) = 0 Here, we have r(x) = p(a) = 0 ∴ (x – 3) is the factor of p(x).

Step-by-step explanation:

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