source: ncert_exercises
topic: algebra
class_level: class_9
chapter: ch2
exercise: ex2.3
difficulty: intermediate
Class 9 | Ch2: Polynomials | Exercise 2.3
Remainder Theorem: If p(x) is divided by (x-a), remainder = p(a).
Q1. Find the remainder when x³+3x²+3x+1 is divided by: (i) x+1 (ii) x-1/2 (iii) x (iv) x+π (v) 5+2x Answer: (i) x+1 → a=-1. p(-1)=(-1)³+3(-1)²+3(-1)+1=-1+3-3+1=0. Remainder=0. ✅ (ii) x-1/2 → a=1/2. p(1/2)=(1/2)³+3(1/2)²+3(1/2)+1=1/8+3/4+3/2+1=27/8. Remainder=27/8. ✅ (iii) x → a=0. p(0)=0+0+0+1=1. Remainder=1. ✅ (iv) x+π → a=-π. p(-π)=(-π)³+3(-π)²+3(-π)+1=-π³+3π²-3π+1. ✅ (v) 5+2x → a=-5/2. p(-5/2)=(-5/2)³+3(-5/2)²+3(-5/2)+1=-125/8+75/4-15/2+1=-27/8. ✅
Q2. Find the remainder when x³-ax²+6x-a is divided by x-a. Answer: a=a (substitute x=a). p(a)=a³-a(a²)+6a-a=a³-a³+6a-a=5a. Remainder=5a. ✅
Q3. Check whether 7+3x is a factor of 3x³+7x. Answer: Zero of 7+3x is x=-7/3. p(-7/3)=3(-7/3)³+7(-7/3)=3(-343/27)+(-49/3)=-343/9-49/3=-343/9-147/9=-490/9≠0. NOT a factor. ✅