Question

Question 2 exercise2.4 class 9th​

Answer 1

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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

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(i) We have, p (x) = 2x³ + x² – 2x – 1 and g (x) = x + 1

∴ p(-1) = 2(-1)³+ (-1)² – 2(-1) – 1

= 2(-1) + 1 + 2 – 1

= -2 + 1 + 2 -1 = 0

⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2

∴ p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1

= -8 + 12 – 6 + 1

= -14 + 13

= -1

⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x³ – 4x² + x + 6 and g (x) = x – 3

∴ p(3) = (3)³ – 4(3)² + 3 + 6

= 27 – 4(9) + 3 + 6

= 27 – 36 + 3 + 6 = 0

⇒ p(3) = 0, so g(x) is a factor of p(x).